To calculate the energy stored in the inductor after 2.60 ms of oscillation, we can use the formula:
f = 1 / (2π√(LC))
Given that the inductance (L) is 90.0 mH and the capacitance (C) is 1.75 nF, we need to convert them to their base units:
L = 90.0 × [tex]10^{(-3)[/tex] H
C = 1.75 × [tex]10^{(-9)[/tex] F
Now we can substitute these values into the formula to find the oscillation frequency:
f = 1 / (2π√(90.0 × [tex]10^{(-3)[/tex] × 1.75 × [tex]10^{(-9)[/tex]))
f ≈ 1 / (2π√(1.575 × [tex]10^{(-11)[/tex])) ≈ 3.189 × [tex]10^7[/tex] Hz
Therefore, the oscillation frequency of the circuit is approximately 3.189 × [tex]10^7[/tex] Hz.
Inductance, L = 90.0 mH = 90.0 × [tex]10^{(-3)[/tex] H
Maximum current, [tex]I_{max[/tex] = 0.810 A
The energy stored in the inductor can be calculated using the formula:
E = 0.5 × L ×[tex]I_{max}^2[/tex]
Substituting the given values:
E = 0.5 × 90.0 × [tex]10^{(-3)[/tex] H × [tex](0.810 A)^2[/tex]
Calculating further:
E ≈ 0.0068 J
Thus, the energy stored in the inductor after 2.60 ms of oscillation is approximately 0.0068 J.
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Required information A 35.0-nC charge is placed at the origin and a 57.0 nC charge is placed on the +x-axis, 2.20 cm from the origin. What is the electric potential at a point halfway between these two charges?
V =
The electric potential at a point halfway between the 35.0 nC charge at the origin and the 57.0 nC charge on the +x-axis is 1.83 kV.
To calculate the electric potential at a point halfway between the two charges, we need to consider the contributions from each charge and sum them together.
Given:
Charge q1 = 35.0 nC at the origin (0, 0).
Charge q2 = 57.0 nC on the +x-axis, 2.20 cm from the origin.
The electric potential due to a point charge at a distance r is given by the formula:
V = k * (q / r),
where V is the electric potential, k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance.
Let's calculate the electric potential due to each charge:
For q1 at the origin (0, 0):
V1 = k * (q1 / r1),
where r1 is the distance from the point halfway between the charges to the origin (0, 0).
For q2 on the +x-axis, 2.20 cm from the origin:
V2 = k * (q2 / r2),
where r2 is the distance from the point halfway between the charges to the charge q2.
Since the point halfway between the charges is equidistant from each charge, r1 = r2.
Now, let's calculate the distances:
r1 = r2 = 2.20 cm / 2 = 1.10 cm = 0.0110 m.
Substituting the values into the formula:
V1 = k * (35.0 x 10^(-9) C) / (0.0110 m),
V2 = k * (57.0 x 10^(-9) C) / (0.0110 m).
Calculating the electric potentials:
V1 ≈ 2863.64 V,
V2 ≈ 4660.18 V.
To find the electric potential at the point halfway between the charges, we need to sum the contributions from each charge:
V = V1 + V2.
Substituting the calculated values:
V ≈ 2863.64 V + 4660.18 V.
Calculating the sum:
V ≈ 7523.82 V.
Therefore, the electric potential at a point halfway between the two charges is approximately 7523.82 volts.
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2. For each pair of systems, circle the one with the larger entropy. If they both have the same entropy, explicitly state it. a. 1 kg of ice or 1 kg of steam b. 1 kg of water at 20°C or 2 kg of water at 20°C c. 1 kg of water at 20°C or 1 kg of water at 50°C d. 1 kg of steam (H₂0) at 200°C or 1 kg of hydrogen and oxygen atoms at 200°C Two students are discussing their answers to the previous question: Student 1: I think that 1 kg of steam and 1 kg of the hydrogen and oxygen atoms that would comprise that steam should have the same entropy because they have the same temperature and amount of stuff. Student 2: But there are three times as many particles moving about with the individual atoms not bound together in a molecule. I think if there are more particles moving, there should be more disorder, meaning its entropy should be higher. Do you agree or disagree with either or both of these students? Briefly explain your reasoning.
a. 1 kg of steam has the larger entropy. b. 2 kg of water at 20°C has the larger entropy. c. 1 kg of water at 50°C has the larger entropy. d. 1 kg of steam (H2O) at 200°C has the larger entropy.
Thus, the answers to the question are:
a. 1 kg of steam has a larger entropy.
b. 2 kg of water at 20°C has a larger entropy.
c. 1 kg of water at 50°C has a larger entropy.
d. 1 kg of steam (H₂0) at 200°C has a larger entropy.
Student 1 thinks that 1 kg of steam and 1 kg of hydrogen and oxygen atoms that make up the steam should have the same entropy because they have the same temperature and amount of stuff. Student 2, on the other hand, thinks that if there are more particles moving around, there should be more disorder, indicating that its entropy should be higher.I agree with student 2's reasoning. Entropy is directly related to the disorder of a system. Higher disorder indicates a higher entropy value, whereas a lower disorder implies a lower entropy value. When there are more particles present in a system, there is a greater probability of disorder, which results in a higher entropy value.
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Problem 1.10 A small spherical ball of mass m and radius R is dropped from rest into a liquid of high viscosity 7, such as honey, tar, or molasses. The only appreciable forces on it are gravity mg and a linear drag force given by Stokes's law, FStokes -6Rv, where v is the ball's velocity, and the minus sign indicates that the drag force is opposite to the direction of v. (a) Find the velocity of the ball as a function of time. Then show that your answer makes sense for (b) small times; (c) large times.
A small spherical ball of mass m and radius R is dropped from rest into a liquid of high viscosity 7, such as honey, tar, or molasses. the velocity is approximately (g/6R), and for large times, the velocity approaches (g/6R) and becomes constant.
(a) To find the velocity of the ball as a function of time, we need to consider the forces acting on the ball. The only two forces are gravity (mg) and the linear drag force (FStokes).
Using Newton's second law, we can write the equation of motion as:
mg - FStokes = ma
Since the drag force is given by FStokes = -6Rv, we can substitute it into the equation:
mg + 6Rv = ma
Simplifying the equation, we have:
ma + 6Rv = mg
Dividing both sides by m, we get:
a + (6R/m) v = g
Since acceleration a is the derivative of velocity v with respect to time t, we can rewrite the equation as a first-order linear ordinary differential equation:
dv/dt + (6R/m) v = g
This is a linear first-order ODE, and we can solve it using the method of integrating factors. The integrating factor is given by e^(kt), where k = 6R/m. Multiplying both sides of the equation by the integrating factor, we have:
e^(6R/m t) dv/dt + (6R/m)e^(6R/m t) v = g e^(6R/m t)
The left side can be simplified using the product rule of differentiation:
(d/dt)(e^(6R/m t) v) = g e^(6R/m t)
Integrating both sides with respect to t, we get:
e^(6R/m t) v = (g/m) ∫e^(6R/m t) dt
Integrating the right side, we have:
e^(6R/m t) v = (g/m) (m/6R) e^(6R/m t) + C
Simplifying, we get:
v = (g/6R) + Ce^(-6R/m t)
where C is the constant of integration.
(b) For small times, t → 0, the exponential term e^(-6R/m t) approaches 1, and we can neglect it. Therefore, the velocity of the ball simplifies to:
v ≈ (g/6R) + C
This means that initially, when the ball is dropped from rest, the velocity is approximately (g/6R), which is constant and independent of time.
(c) For large times, t → ∞, the exponential term e^(-6R/m t) approaches 0, and we can neglect it. Therefore, the velocity of the ball simplifies to:
v ≈ (g/6R)
This means that at large times, when the ball reaches a steady-state motion, the velocity is constant and equal to (g/6R), which is determined solely by the gravitational force and the drag force.
In summary, the velocity of the ball as a function of time is given by:
v = (g/6R) + Ce^(-6R/m t)
For small times, the velocity is approximately (g/6R), and for large times, the velocity approaches (g/6R) and becomes constant.
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(a) the energy released per event in joules ] (b) the change in mass (in kg ) during the event ×kg [0/1.92 Points] SERCP11 30.4.OP.021. In a pair-production reaction, a photon produces a muon-antimuon pair. γ→μ −
+μ +
The rest energy of a muon is 105.7MeV. (a) What is the lowest possible frequency (in Hz ) of the photon that can produce the muon-antimuon pair? Hz (b) What is the wavelength (in m ) that corresponds to this lowest possible frequency? 2s What is the relationship between frequency, wavelength, and the speed of light? m
Lowest possible frequency: 4.84 x 10^20 Hz, Corresponding wavelength: 6.19 x 10^-13 m (or 2s), The relationship between frequency, wavelength, and the speed of light is given by c = fλ.
The lowest possible frequency (f) of the photon that can produce the muon-antimuon pair can be found by using the equation E = hf, where E is the energy (rest energy of the muon in this case) and h is the Planck's constant (approximately 6.63 x 10^-34 J·s). Converting the rest energy of the muon from MeV to joules (1 MeV = 1.6 x 10^-13 J), we have E = 105.7 MeV = 105.7 x 1.6 x 10^-13 J. By rearranging the equation, we can solve for the frequency: f = E / h. Plugging in the values, we get f = (105.7 x 1.6 x 10^-13 J) / (6.63 x 10^-34 J·s) ≈ 4.84 x 10^20 Hz. (b) The relationship between frequency (f), wavelength (λ), and the speed of light (c) is given by the equation c = fλ, where c is the speed of light (approximately 3 x 10^8 m/s). Rearranging the equation, we can solve for the wavelength: λ = c / f. Plugging in the values, we get λ = (3 x 10^8 m/s) / (4.84 x 10^20 Hz) ≈ 6.19 x 10^-13 m or 2s (as mentioned in the question).
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The magnetic force on a straight wire 0.30 m long is 2.6 x 10^-3 N. The current in the wire is 15.0 A. What is the magnitude of the magnetic field that is perpendicular to the wire?
Answer: the magnitude of the magnetic field perpendicular to the wire is approximately 1.93 x 10^-3 T.
Explanation:
The magnetic force on a straight wire carrying current is given by the formula:
F = B * I * L * sin(theta),
where F is the magnetic force, B is the magnetic field, I is the current, L is the length of the wire, and theta is the angle between the magnetic field and the wire (which is 90 degrees in this case since the field is perpendicular to the wire).
Given:
Length of the wire (L) = 0.30 m
Current (I) = 15.0 A
Magnetic force (F) = 2.6 x 10^-3 N
Theta (angle) = 90 degrees
We can rearrange the formula to solve for the magnetic field (B):
B = F / (I * L * sin(theta))
Plugging in the given values:
B = (2.6 x 10^-3 N) / (15.0 A * 0.30 m * sin(90 degrees))
Since sin(90 degrees) equals 1:
B = (2.6 x 10^-3 N) / (15.0 A * 0.30 m * 1)
B = 2.6 x 10^-3 N / (4.5 A * 0.30 m)
B = 2.6 x 10^-3 N / 1.35 A*m
B ≈ 1.93 x 10^-3 T (Tesla)
A delivery truck travels 31 blocks north, 18 blocks east, and 26 blocks south. Assume the blooks are equal length What is the magnitude of its final displacement from the origin? What is the direction of its final displacement from the origin? Express your answer using two significant figures.
The magnitude of final displacement is from the origin is approximately 36 blocks and the direction of the final displacement from the origin is approximately 59° (measured counterclockwise from the positive x-axis or east direction).
To calculate the magnitude of the final displacement, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, we can consider the north-south displacement as one side and the east-west displacement as the other side of a right triangle. The final displacement is the hypotenuse of this triangle.
Given:
North displacement = 31 blocks (positive value)
East displacement = 18 blocks (positive value)
South displacement = 26 blocks (negative value)
To calculate the magnitude of the final displacement:
Magnitude = sqrt((North displacement)^2 + (East displacement)^2)
Magnitude = sqrt((31)^2 + (18)^2)
Magnitude = sqrt(961 + 324)
Magnitude = sqrt(1285)
Magnitude ≈ 35.88
Rounded to two significant figures, the magnitude of the final displacement from the origin is approximately 36 blocks.
To determine the direction of the final displacement from the origin, we can use trigonometry. We can calculate the angle with respect to a reference direction, such as north or east.
Angle = atan((North displacement) / (East displacement))
Angle = atan(31 / 18)
Angle ≈ 59.06°
Rounded to two significant figures, the direction of the final displacement from the origin is approximately 59° (measured counterclockwise from the positive x-axis or east direction).
Thus, rounded to two significant figures, the magnitude of final displacement is from the origin is approximately 36 blocks and the direction of the final displacement from the origin is approximately 59° (measured counterclockwise from the positive x-axis or east direction).
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A sculpture weighing 35000 N rests on a horizontal surface at the top of a 1.8 m high stand (Figure 2). The stand's cross-sectional area is 7.3 x 102 m2 and it is made of granite with a
Young's modulus of 4.5 x 1010 Pa. By how much does the sculpture compress the stand?
[3]
Figure 2
A. 1.9 x 10-2 mm
B. 5.2 x 102 mm
C. 32.85 x 10-2 mm
D. 6.3 x 102 mm
The sculpture compresses the stand by correct option A) 1.9 x 10-2 mm. Compression can be determined by dividing the applied force by the product of the cross-sectional area and the material's Young's modulus.
To calculate the compression of the stand, we can use Hooke's Law, which states that the deformation of a material is directly proportional to the applied force and inversely proportional to its stiffness or Young's modulus.
The weight of the sculpture is 35000 N, and it applies a force on the stand. This force causes the stand to compress.
Using the formula for compression, Δx = F/(A * E), where Δx is the compression, F is the force, A is the cross-sectional area, and E is the Young's modulus of the material, we can calculate the compression of the stand.
Δx = (35000 N) / ((7.3 x [tex]10^{2}[/tex] [tex]m^{2}[/tex]) * (4.5 x [tex]10^{10}[/tex] Pa))
Simplifying the expression, we find that the sculpture compresses the stand by approximately 1.9 x [tex]10^{-2}[/tex] mm.
Therefore, the correct answer is A. 1.9 x 10-2 mm.
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A copper calorimetric cup with a mass of 100g contains 96g of water at 13 C. If 70g of a substance at 84 degC is dropped into the cup, the temperature increases to 20 degC. Find the specific heat capacity of the substance.
Someone pours 150g of heated lead shot into a 250g aluminum calorimeter cup that contains 200g of water at 25 degC . The final temperature is 28 degC. What was the intial temperature of the lead shot?
What mass of water at 50 degC can be converted into steam at 110 degC by 9.6 x10^6 J?
Answer: The mass of water required is 4247.79 g (answer).
Therefore, the mass of water at 50°C that can be converted into steam at 110°C by 9.6 × 106 J is 4247.79 g.
Question 1 : A copper calorimetric cup with a mass of 100g contains 96g of water at 13 C. If 70g of a substance at 84 degC is dropped into the cup, the temperature increases to 20 degC. Find the specific heat capacity of the substance.
Solution :The amount of heat lost by hot body = amount of heat gained by cold body
Applying the formula of specific heat capacity
mcΔT = msΔT
Since there is no loss of heat to the surrounding mcΔT = msΔT
m1c1ΔT1 = m2s2ΔT2
where m1, c1 and ΔT1 are the mass, specific heat capacity and the temperature change of the copper cup and water.
m2, s2 and ΔT2 are the mass, specific heat capacity and the temperature change of the substance.
We know that the mass of copper calorimetric cup = 100g
the mass of water = 96g
the temperature of water = 13°C
the mass of the substance = 70g
the temperature of the substance = 84°C
The final temperature after mixing = 20°C
Temperature change of the substance,
ΔT2 = Final temperature - initial temperature
= 20°C - 84°C= - 64°C
Temperature change of the water,
ΔT1 = Final temperature - initial temperature
= 20°C - 13°C= 7°C
Thus, by substituting the values in the formula:
m1c1ΔT1 = m2s2ΔT2(100 g) (0.385 J/g°C) (7°C)
= (70 g) s2 (-64°C)s2
= 0.448 J/g°C
Specific heat capacity of the substance is 0.448 J/g°C (answer)
Hence, the specific heat capacity of the substance is 0.448 J/g°C.
Question 2: Someone pours 150g of heated lead shot into a 250g aluminum calorimeter cup that contains 200g of water at 25°C. The final temperature is 28°C. What was the initial temperature of the lead shot?
Solution:
Heat lost by lead shot = Heat gained by water + Heat gained by Aluminium container Q1 = Q2 + Q3
The formula of heat: Q = m × c × ΔT
Where,Q1 = Heat lost by lead shot
m = mass of the object
c = Specific heat capacity
ΔT = Temperature difference.
Q2 = Heat gained by water
m = mass of the object
c = Specific heat capacity
ΔT = Temperature difference.
Q3 = Heat gained by Aluminium container
m = mass of the object
c = Specific heat capacity
ΔT = Temperature difference.
Substitute the values given in the question,Q1 = (150 g) × c × (Ti - 28) °C
Q2 = (200 g) × 4.18 J/g°C × (28 - 25) °C
= 2502 JQ3 = (250 g) × 0.897 J/g°C × (28 - 25) °C
= 672.75 J Q1 = Q2 + Q3(150 g) × c × (Ti - 28) °C
= 2502 J + 672.75 J(150 g) × c × (Ti - 28) °C
= 3174.75 J(150 g) × c × (Ti - 28) / 150 g
= 3174.75 J / 150 gTi - 28
= 21.16°C (Approx.)Ti
= 49.16°C (answer)
Hence, the initial temperature of the lead shot was 49.16°C.
Question 3 : What mass of water at 50°C can be converted into steam at 110°C by 9.6 x 10^6 J?
Solution:
To find the mass of water, we use the formula, Q = mL
Where,
Q = Amount of heat required to change the phase of water from liquid to gas
L = Latent heat of vaporisation
m = Mass of water required.
To find the value of L, we use the specific heat capacity of water.The amount of heat required to raise 1 g of water by 1°C = 1 cal/g°C
Specific heat capacity of water = 4.18 J/g°C
Amount of heat required to raise 1 g of water by 1°C = 4.18 J/g°C
Specific latent heat of vaporisation of water = 2260 J/g
Amount of heat required to convert 1 g of water into steam = 2260 J/g
To find the mass of water,m = Q / LWhere,
Q = 9.6 × 106 J (Given)
L = 2260 J/g
Substitute the given values in the formula,
m = 9.6 × 106 J / 2260 J/g
m = 4247.79 g (Approx.)
Hence, the mass of water required is 4247.79 g (answer).
Therefore, the mass of water at 50°C that can be converted into steam at 110°C by 9.6 × 106 J is 4247.79 g.
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Consider the following potential: Voi x≤0 V(x) = {-Vo; 0 < x a Assuming that the flux of particles are incident from the right, and the energy of the particles are 0 < E< Vo, find the amplitude of the reflected wave in the region > a
The amplitude of the reflected wave in the region x > a is given by Ar = -Ai √(1 - E / Vo) e^(-ik1a).
The given potential is a step potential of height -Vo at x ≤ 0, and 0 at 0 < x < a, and height 0 beyond x > a.
The probability current density J for a particle of energy E in a given region is given as J = (h / 2πi) [ψ*(dψ / dx) - (dψ* / dx) ψ]where ψ is the wave function and ψ* is its complex conjugate.
Using the probability current density expression, we can write down the transmission and reflection coefficients. The transmission coefficient T is the probability flux transmitted through the barrier, and the reflection coefficient R is the probability flux reflected from the barrier. The probability flux J is proportional to the square of the amplitude of the wave. Thus, we can write the transmission and reflection coefficients as:
T = |At|² / |Ai|² and R = |Ar|² / |Ai|²
where At is the amplitude of the transmitted wave, Ar is the amplitude of the reflected wave, and Ai is the amplitude of the incident wave.
Now, let's solve the problem at hand.
A particle of energy E is incident from the right, with an amplitude of Ai. The wave function for the particle in the region x ≤ 0 is given as:
ψ1(x) = Ae^(ik1x) + Be^(-ik1x), where k1 = √(2m(E + Vo)) / h and A and B are constants.
The wave function for the particle in the region 0 < x < a is given as:
ψ2(x) = Ce^(ik2x) + De^(-ik2x), where k2 = √(2mE) / h and C and D are constants.
The wave function for the particle in the region x > a is given as:
ψ3(x) = Ee^(ik3x), where k3 = √(2mE) / h and E is a constant.
Note that we have assumed that the potential is zero in the region x > a.
Using the boundary conditions at x = 0 and x = a, we can solve for the constants A, B, C, D, and E in terms of Ai as follows:
A = Ai / 2 + Ar / 2, B = Ai / 2 - Ar / 2, C = Ae^(ik1a) + Be^(-ik1a), D = Ae^(-ik1a) + Be^(ik1a), and E = Ce^(ik2a).
Now, we can calculate the reflection and transmission coefficients as:
R = |Ar|² / |Ai|² = |B - Ai / 2|² / |Ai|² = |Ai / 2 - (Ai / 2) e^(-2ik1a)|² / |Ai|² = |1/2 - 1/2 e^(-2ik1a)|² = sin²(k1a)T = |At|² / |Ai|² = |E|² / |Ai|² = |Ce^(ik2a)|² / |Ai|² = |C|² / |Ai|² = 1 - sin²(k1a)
Thus, we have derived the reflection and transmission coefficients in terms of the incident amplitude Ai and the energy E of the particle. For particles with energy 0 < E < Vo, we have sin(k1a) = √(1 - E / Vo) and cos(k1a) = √(E / Vo). The amplitude of the reflected wave in the region x > a is given by Ar = -Ai / 2 e^(-ik1a) (1 - e^(-2ik1a)).Thus, we have Ar = -Ai sin(k1a) e^(-ik1a).
Hence, the amplitude of the reflected wave in the region x > a is given by Ar = -Ai √(1 - E / Vo) e^(-ik1a).
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Light of wavelength 648.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 84.5 cm from the slit. The distance on the screen between the fourth order minimum and the central maximum is 1.93 cm . What is the width of the slit in micrometers (μm)?
= μm
The width of the slit is determined to be in micrometers (μm).The width of the slit can be determined using the formula for the slit diffraction pattern. In this case, we are given the wavelength of light (648.0 nm), the distance from the slit to the screen (84.5 cm), and the distance on the screen between the fourth order minimum and the central maximum (1.93 cm).
The width of the slit can be calculated using the equation d*sin(theta) = m*lambda, where d is the width of the slit, theta is the angle of diffraction, m is the order of the minimum, and lambda is the wavelength of light.
First, we need to find the angle of diffraction for the fourth order minimum. We can use the small angle approximation, which states that sin(theta) ≈ tan(theta) ≈ y/L, where y is the distance on the screen and L is the distance from the slit to the screen.
Using the given values, we can calculate the angle of diffraction for the fourth order minimum. Then, we can rearrange the equation to solve for the slit width d.
After performing the necessary calculations, the widwidth of the slit is determined to be in micrometers (μm).
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Determine the upward force that the biceps muscle exerts when a 75 Newton load is held in the hand when the arm is at 900 angles as shown. If the combined weight of the forearm and hand is assumed to be 35 Newton’s and acts at the center of gravity.
The total upward force exerted by the biceps muscle when holding the 75 Newton load in the hand at a 90-degree angle is 110 Newtons
To determine the upward force exerted by the biceps muscle when holding a 75 Newton load in the hand at a 90-degree angle, we need to consider the forces acting on the arm. The total force exerted by the biceps muscle can be calculated by summing the upward force required to counteract the load's weight and the weight of the forearm and hand. Given that the combined weight of the forearm and hand is 35 Newtons and acts at the center of gravity, the force required to counteract this weight is 35 Newtons in the downward direction. To maintain equilibrium, the biceps muscle must exert an equal and opposite force of 35 Newtons in the upward direction. Additionally, since the load in the hand weighs 75 Newtons, the biceps muscle needs to exert an additional 75 Newtons in the upward direction to counteract its weight. Therefore, the total upward force exerted by the biceps muscle when holding the 75 Newton load in the hand at a 90-degree angle is 110 Newtons.
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Q3. A hanging platform has four cylindrical supporting cables of diameter 2.5 cm. The supports are made from solid aluminium, which has a Young's Modulus of Y = 69 GPa. The weight of any object placed on the platform is equally distributed to all four cables. a) When a heavy object is placed on the platform, the cables are extended in length by 0.4%. Find the mass of this object. (3) b) Poisson's Ratio for aluminium is v= 0.33. Calculate the new diameter of the cables when supporting this heavy object. (3) (6 marks)
The new diameter of the cable is 0.892 cm. Option (ii) is the correct answer.
Given: Diameter of supporting cables,
d = 2.5 cm Young's Modulus of aluminium,
Y = 69 GPa Load applied,
F = mg
Extension in the length of the cables,
δl = 0.4% = 0.004
a) Mass of the object placed on the platform can be calculated as:
m = F/g
From the question, we know that the weight of any object placed on the platform is equally distributed to all four cables.
So, weight supported by each cable = F/4
Extension in length of each cable = δl/4
Young's Modulus can be defined as the ratio of stress to strain.
Y = stress/strainstress = Force/areastrain = Extension in length/Original length
Hence, stress = F/4 / (π/4) d2 = F/(π d2)strain = δl/4 / L
Using Hooke's Law, stress/strain
= Yπ d2/F = Y δl/Ld2 = F/(Y δl/π L) = m g / (Y δl/π L)
On substituting the given values, we get:
d2 = (m × 9.8) / ((69 × 10^9) × (0.004/100) / (π × 2.5/100))d2 = 7.962 × 10^-5 m2
New diameter of the cable is:
d = √d2 = √(7.962 × 10^-5) = 0.00892 m = 0.892 cm
Therefore, the new diameter of the cable is 0.892 cm.
Hence, option (ii) is the correct answer.
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8) If the refracting index of light in a medium is n = 2.7, what is the speed of light in the medium? Find the wavelength of an EM wave with a frequency of 12 x 10° Hz in the medium with n = 2.7.
The speed of light in the medium with a refractive index of 2.7 is approximately 1.11 x 10⁸ meters per second. The wavelength of the EM wave is approximately 9.25 meters.
The speed of light in a medium can be calculated using the formula v = c/n, where v is the speed of light in the medium, c is the speed of light in a vacuum, and n is the refractive index of the medium.
In this case, the refractive index of the medium is given as n = 2.7. The speed of light in a vacuum is approximately 3 x 10⁸ meters per second.
Plugging these values into the formula, we get
v = (3 x 10⁸ m/s) / 2.7. Simplifying this expression gives us v ≈ 1.11 x 10^8 meters per second.
Therefore, the speed of light in the medium with a refractive index of 2.7 is approximately 1.11 x 10⁸ meters per second.
To find the wavelength of an electromagnetic wave with a frequency of 12 x 10⁶ Hz in the medium with n = 2.7, we can use the formula λ = v/f, where λ is the wavelength, v is the speed of light in the medium, and f is the frequency of the wave.
Using the previously calculated speed of light in the medium (v = 1.11 x 10⁸ m/s) and the given frequency (f = 12 x 10⁶ Hz), we can calculate the wavelength:
λ = (1.11 x 10⁸ m/s) / (12 x 10⁶ Hz) ≈ 9.25 meters.
Therefore, the wavelength of the EM wave with a frequency of 12 x 10⁶ Hz in the medium with n = 2.7 is approximately 9.25 meters.
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6. GO A plate carries a charge of 3.0 uC, while a rod carries a charge of +2.0 uC. How many electrons must be transferred from the plate to the rod, so that both objects have the same charge?
Approximately 6.24 x 10¹² electrons must be transferred from the plate to the rod for both objects to have the same charge.
To determine the number of electrons that must be transferred from the plate to the rod, we need to consider the elementary charge and the difference in charge between the two objects.
The elementary charge is the charge carried by a single electron, which is approximately 1.602 x 10⁻¹⁹ coulombs (C). The charge carried by an electron is approximately -1.602 x 10⁻¹⁹ coulombs (C).
Given that the plate carries a charge of 3.0 μC (microcoulombs) and the rod carries a charge of +2.0 μC, we need to find the difference in charge between them.
Converting the charges to coulombs:
Plate charge = 3.0 μC = 3.0 x 10⁻⁶ C
Rod charge = +2.0 μC = 2.0 x 10⁻⁶ C
The difference in charge is:
Difference in charge = Plate charge - Rod charge
= 3.0 x 10⁻⁶ C - 2.0 x 10⁻⁶ C
= 1.0 x 10⁻⁶ C
Since the plate has an excess of charge, electrons need to be transferred to the rod, which has a positive charge. The charge of an electron is -1.602 x 10^-19 C, so the number of electrons transferred can be calculated as:
Number of electrons transferred = Difference in charge / Charge of an electron
= 1.0 x 10⁻⁶ C / (1.602 x 10⁻¹⁹ C)
≈ 6.24 x 10¹² electrons
Therefore, approximately 6.24 x 10¹² electrons must be transferred from the plate to the rod for both objects to have the same charge.
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1. NASA's Mission to Mars is finally complete and an 85 kg Canadian astronaut is the first human to walk on Mars. If Mars has a mass of 6.37 x 10²3 kg and a radius of 3.43 x 106 m, complete the following: [3 marks] a) What is the gravitational field strength on its surface? [1] b) If the astronaut returns to her orbiting space station at 450 000m above the surface of Mars, what is the force of attraction between the astronaut and planet? [2]\
a) Calculation of Gravitational field strength Gravitational field strength is the force exerted per unit mass. It is a vector quantity and it is denoted by g.
It is expressed in units of N/kg.
Using the formula, g = GM/r²Where,G = Universal gravitational constant = 6.67 x 10-11 Nm²/kg²M = Mass of the planet = 6.37 x 1023 kgr = Radius of the planet = 3.43 x 106 m
Substituting the values in the above formula,g = (6.67 x 10-11) x (6.37 x 1023) / (3.43 x 106)² = 3.71 N/kg
Hence, the gravitational field strength on Mars is 3.71 N/kg.b)
Calculation of Force of attraction between astronaut and planetUsing the formula F = (GmM)/r²Where,G = Universal gravitational constant = 6.67 x 10-11 Nm²/kg²m = Mass of the astronaut = 85 kgM = Mass of the planet = 6.37 x 1023 kgr = Distance between the astronaut and the planet = 3.43 x 106 + 450000 = 3.88 x 106 m
Substituting the values in the above formula,F = (6.67 x 10-11 x 85 x 6.37 x 1023)/ (3.88 x 106)² = 780 N (approx)
Therefore, the force of attraction between the astronaut and planet is 780 N (approx).
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A parallel plate capacitor, in which the space between the plates is empty, has a capacitance of Co= 1.5μF and it is connected to a battery whose voltage is V = 2.7V and fully charged. Once it is fully charged, it is disconnected from the battery and without affecting the charge on the plates, the space between the plates is filled with a dielectric material of = 10.7. How much change occurs in the energy of the capacitor (final energy minus initial energy)? Express your answer in units of mJ (mili joules) using two decimal places.
The change in energy of the capacitor is 51.93 μJ (microjoules), which can be expressed as 0.05193 mJ (millijoules) when rounded to two decimal places.
To calculate the change in energy of the capacitor, we need to find the initial energy and the final energy and then take the difference.
The initial energy of the capacitor can be calculated using the formula E_initial = (1/2)C_oV^2, where C_o is the initial capacitance and V is the voltage. In this case, C_o = 1.5 μF and V = 2.7V. Plugging in these values, we get E_initial = (1/2)(1.5 μF)(2.7V)^2.
So, Initial energy, E_initial = (1/2)C_oV^2
Substituting C_o = 1.5 μF and V = 2.7V:
E_initial = (1/2)(1.5 μF)(2.7V)^2
E_initial = 6.1575 μJ (microjoules)
After the space between the plates is filled with a dielectric material, the capacitance changes. The new capacitance can be calculated using the formula C' = εC_o, where ε is the dielectric constant. In this case, ε = 10.7. Therefore, the new capacitance is C' = 10.7(1.5 μF).
So, New capacitance, C' = εC_o
Substituting ε = 10.7 and C_o = 1.5 μF:
C' = 10.7(1.5 μF)
C' = 16.05 μF
The final energy of the capacitor can be calculated using the formula E_final = (1/2)C'V^2, where C' is the new capacitance and V is the voltage. Plugging in the values, we get E_final = (1/2)(10.7)(1.5 μF)(2.7V)^2.
So, Final energy, E_final = (1/2)C'V^2
Substituting C' = 16.05 μF and V = 2.7V:
E_final = (1/2)(16.05 μF)(2.7V)^2
E_final = 58.0833 μJ (microjoules)
To find the change in energy, we subtract the initial energy from the final energy: ΔE = E_final - E_initial.
Therefore, Change in energy (ΔE):
ΔE = E_final - E_initial
ΔE = 58.0833 μJ - 6.1575 μJ
ΔE = 51.9258 μJ (microjoules)
So, the energy change is 51.9258 μJ or 0.05193 mJ.
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A circuit consists of an 110- resistor in series with a 5.0-μF capacitor, the two being connected between the terminals of an ac generator. The voltage of the generator is fixed. At what frequency is the current in the circuit one-half the value that exists when the frequency is very large? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise
The peak value of the current supplied by the generator is approximately 2.07 Amperes.
To determine the peak value of the current supplied by the generator, we can use the relationship between voltage, current, and inductance in an AC circuit.
The peak current (I_peak) can be calculated using the formula:
I_peak = V_rms / (ω * L),
where:
V_rms is the root mean square (RMS) value of the voltage (in this case, 9.0 V),
ω is the angular frequency of the AC signal (in radians per second), and
L is the inductance of the inductor (in henries).
To convert the given frequency (690 Hz) to angular frequency (ω), we can use the formula:
ω = 2πf,
where:
f is the frequency.
Substituting the values into the formula, we have:
ω = 2π * 690 Hz ≈ 4,335.48 rad/s.
Now, let's calculate the peak current:
I_peak = (9.0 V) / (4,335.48 rad/s * 10 × 10^(-3) H).
Simplifying the expression:
I_peak ≈ 2.07 A.
Therefore, the peak value of the current supplied by the generator is approximately 2.07 Amperes.
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Make a a derivation for the unknown resistor equation (Rx) in
terms of voltages and lengths on the wheatstone bridge
The unknown resistor (Rx) in a Wheatstone bridge circuit can be determined using the equation:
Rx = (V_out1 * R2) / (V_in - V_out2)
This equation relates Rx to the voltages V_out1 and V_out2, as well as the resistance R2 and the input voltage V_in.
Let's consider a typical Wheatstone bridge circuit consisting of four resistors: R1, R2, R3, and Rx. The bridge is supplied with a known voltage V_in and has two outputs: V_out1 and V_out2.
1. First, let's find the relationship between the voltages V_out1 and V_out2 in terms of the resistors. According to Kirchhoff's voltage law, the voltage drop across any closed loop in a circuit is zero. Applying this law to the two loops in the Wheatstone bridge, we have:
Loop 1: V_in = V_out1 + I1 * R1 + I2 * Rx
Loop 2: V_in = V_out2 + I3 * R3 + I2 * (R2 + Rx)
Where I1, I2, and I3 are the currents flowing through R1, Rx, and R3, respectively.
2. To simplify the equations, we can express I1, I2, and I3 in terms of the voltages and resistances using Ohm's law. Assuming the resistors have negligible internal resistance, we have:
I1 = V_out1 / R1
I2 = (V_out1 - V_out2) / (R2 + Rx)
I3 = V_out2 / R3
Substituting these values back into the loop equations, we get:
V_in = V_out1 + (V_out1 - V_out2) * Rx / (R2 + Rx)
V_in = V_out2 + V_out2 * R2 / (R2 + Rx)
3. Now, we can solve these two equations simultaneously to eliminate V_out1 and V_out2. Multiplying the first equation by (R2 + Rx) and the second equation by Rx, we get:
V_in * (R2 + Rx) = V_out1 * (R2 + Rx) + (V_out1 - V_out2) * Rx
V_in * Rx = V_out2 * Rx + V_out2 * R2
4. By rearranging these equations, we can isolate Rx:
V_in * Rx - V_out2 * Rx = V_out1 * (R2 + Rx) - (V_out1 - V_out2) * Rx
V_in * Rx - V_out2 * Rx = V_out1 * R2 + V_out1 * Rx - V_out1 * Rx + V_out2 * Rx
V_in * Rx - V_out2 * Rx = V_out1 * R2 + V_out2 * Rx
Rx * (V_in - V_out2) = V_out1 * R2
Rx = (V_out1 * R2) / (V_in - V_out2)
Therefore, the equation for the unknown resistor Rx in terms of the voltages and lengths on the Wheatstone bridge is:
Rx = (V_out1 * R2) / (V_in - V_out2)
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"Two converging lenses with the same focal length of 10 cm are 40
cm apart. If an object is located 15 cm from one of the lenses,
find the final image distance of the object.
a. 0 cm
b. 5 cm
c. 10 cm
d 15 cm
The final image distance of the object, if the object is located 15 cm from one of the lenses is 6 cm. So none of the options are correct.
To determine the final image distance of the object in the given setup of two converging lenses, we can use the lens formula:
1/f = 1/di - 1/do
Where: f is the focal length of the lens, di is the image distance, do is the object distance.
Given that both lenses have the same focal length of 10 cm, we can consider them as a single lens with an effective focal length of 10 cm. The lenses are 40 cm apart, and the object distance (do) is 15 cm.
Using the lens formula, we can rearrange it to solve for di:
1/di = 1/f + 1/do
1/di = 1/10 cm + 1/15 cm
= (15 + 10) / (10 * 15) cm⁻¹
= 25 / 150 cm⁻¹
= 1 / 6 cm⁻¹
di = 1 / (1 / 6 cm⁻¹) = 6 cm
Therefore, the final image distance of the object is 6 cm. So, none of the options are correct.
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The voltage and Power ratings of a Particular light bulb, which are It's normal operating values are lov and 60w. Assume the resistance of the of with ating Conditions. If the light bulb is operated with a Current that is 50% of the current. rating Idrawn by the bulb? of the bulb, what is the actual Power
The actual power consumed is 30W when the light bulb is worked with a current that is 50% of its current rating using Ohm's Law.
Normal operating value = 60W
Bulb operation = 50% of current.
The relation between voltage, current, and resistance is given by Ohm's Law.
V = I * R.
R = V / I
The formula used for calculating the power rating in normal operating conditions is:
P_0 = V_0 * I_0
The actual current drawn by the bulb I_actual is:
V_0 = I_actual * R
R = V_0 / I_actual
P_actual = V_0 * I_actual
Substituting the values we get:
P_actual = V_0 * I_actual = V_0 * (0.5 * I_0)
60W = V_0 * I_0
V_0 = 60W / I_0
P_actual = (60W / I_0) * (0.5 * I_0) = 30W
Therefore, we can conclude that the actual power consumed is 30W.
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please write a full paraphrasing for the text below. thanks
The magnitude of each of the electric forces with which two point charges at rest interact is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them and has the direction of the line that joins them. . The force is repulsive if the charges are of the same sign, and attractive if they are of the opposite sign. Coulomb's law does comply with the principles of superposition since it determines the electric force of attraction or repulsion experienced by a point charge in the presence of another. The electrical forces between two charges can vary since in some the charges or the distance between them are doubled.
The text states Coulomb's law which expresses that the magnitude of electric forces between two point charges, which are stationary, is proportional to both charges' magnitudes and inversely proportional to the distance square between them.
If two point charges are in the same direction, they repel, and if they are in opposite directions, they attract.Coulomb's law follows the superposition concept, which calculates the repulsion or attraction electric force between a point charge in the presence of another point charge. Due to the doubled distance or charges, the electrical forces between two charges may differ.
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In a Photoelectric effect experiment, the incident photons each has an energy of 4.713×10 −19 J. The power of the incident light is 0.9 W. (power = energy/time) The work function of metal surface used is W 0 = 2.71eV. 1 electron volt (eV)=1.6×10 −19 J. If needed, use h=6.626×10 −34 J⋅s for Planck's constant and c=3.00×10 8 m/s for the speed of light in a vacuum. Part A - How many photons in the incident light hit the metal surface in 7.0 s ? Part B - What is the max kinetic energy of the photoelectrons? Part C - Use classical physics fomula for kinetic energy, calculate the maximum speed of the photoelectrons. The mass of an electron is 9.11×10 −31 kg
The incident photons energy is 1.337 × 10²². The max kinetic energy of the photoelectrons is 6.938 × 10⁻¹ eV. The maximum speed of the photoelectrons is 5.47 × 10⁵ m/s. The correct answer for a) 1.337 × 10²² photons b) 6.938 × 10⁻¹ eV c) 5.47 × 10⁵ m/s
Part A The power of the incident light, P = 0.9 W Total energy delivered, E = P x tE = 0.9 x 7 = 6.3 JThe energy of each photon, E = 4.713 × 10⁻¹⁹ J Number of photons, n = E/E = 6.3/4.713 × 10⁻¹⁹ = 1.337 × 10²² photons
Part B The energy of a photon = hν, where ν is the frequencyν = c/λ where c = speed of light and λ is the wavelength of light.λ = hc/E = hc/ (4.713 × 10⁻¹⁹) = 1.324 × 10⁻⁷ m Kinetic energy of a photoelectron is given by KE max = hν - W₀ = hc/λ - W₀ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸)/1.324 × 10⁻⁷ - (2.71 × 1.6 × 10⁻¹⁹) = 1.11 × 10⁻¹⁹ J = 6.938 × 10⁻¹ eV
Part C Maximum speed of a photoelectron can be calculated by using classical mechanics equation: KEmax = (1/2)mv²where m is the mass of electron and v is the maximum speed. Rearranging gives: v = √(2KEmax/m) = √(2(6.938 × 10⁻¹ eV)(1.6 × 10⁻¹⁹ J/eV)/(9.11 × 10⁻³¹ kg)) = 5.47 × 10⁵ m/s (to 3 significant figures) Answer:a) 1.337 × 10²² photonsb) 6.938 × 10⁻¹ eVc) 5.47 × 10⁵ m/s
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A ray of light traveling in water is incident on an interface with a flat piece of glass. The wavelength of the light in the water is 722 nm and its wavelength in the glass is 543 nm. If the ray in water makes an angle of 45.0 ∘
with respect to the normal to the interface, what angle does the refracted ray in the glass make with respect to the normal?
The refracted ray in the glass makes an angle of approximately 48.4° with respect to the normal.
To determine the angle of the refracted ray in the glass, we can use Snell's Law, which relates the angles and indices of refraction of light as it passes through different mediums. Snell's Law states that the ratio of the sines of the angles of incidence (θ₁) and refraction (θ₂) is equal to the ratio of the indices of refraction (n₁ and n₂) of the two mediums.
In this case, the incident angle in water (θ₁) is given as 45.0°, the wavelength of light in water (λ₁) is 722 nm, and the wavelength of light in glass (λ₂) is 543 nm.
We know that the index of refraction (n) of a medium is inversely proportional to the wavelength of light passing through it, so we can use the ratio of the wavelengths to calculate the ratio of the indices of refraction:
n₁ / n₂ = λ₂ / λ₁
Substituting the given values, we have:
n₁ / n₂ = 543 nm / 722 nm
To simplify the calculation, we can convert the wavelengths to meters:
n₁ / n₂ = (543 nm / 1) / (722 nm / 1) = 0.751
Now, we can apply Snell's Law:
sin(θ₁) / sin(θ₂) = n₂ / n₁
sin(θ₂) = (n₁ / n₂) * sin(θ₁)
Plugging in the values, we get:
sin(θ₂) = 0.751 * sin(45.0°)
To find the angle θ₂, we can take the inverse sine (or arcsine) of both sides:
θ₂ = arcsin(0.751 * sin(45.0°))
Evaluating this expression, we find:
θ₂ ≈ 48.4°
Therefore, the refracted ray in the glass makes an angle of approximately 48.4° with respect to the normal.
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X-rays with an energy of 339 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 57.7^{\circ}∘relative to the incident X-rays, what is the wavelength of the scattered photon?
Answer:
The
wavelength
of the scattered photon is approximately 1.11 × 10^(-11) meters.
Explanation:
Compton scattering is a phenomenon where X-rays interact with electrons, resulting in a shift in wavelength. To determine the wavelength of the scattered photon, we can use the Compton scattering formula:
Δλ = λ' - λ = λ_c * (1 - cos(θ))
Where:
Δλ is the change in wavelength
λ' is the wavelength of the scattered photon
λ is the wavelength of the incident X-ray photon
λ_c is the Compton wavelength (approximately 2.43 × 10^(-12) m)
θ is the scattering angle
Given:
Energy of the incident X-ray photon (E) = 339 keV = 339 * 10^3 eV
Scattering angle (θ) = 57.7 degrees
First, let's calculate the wavelength of the incident X-ray photon using the energy-wavelength relationship:
E = hc / λ
Where:
h is Planck's constant (approximately 6.63 × 10^(-34) J·s)
c is the speed of light (approximately 3.00 × 10^8 m/s)
Converting the energy to joules:
E = 339 * 10^3 eV * (1.60 × 10^(-19) J/eV) = 5.424 × 10^(-14) J
Rearranging the equation to solve for λ:
λ = hc / E
Substituting the values:
λ = (6.63 × 10^(-34) J·s * 3.00 × 10^8 m/s) / (5.424 × 10^(-14) J) ≈ 1.22 × 10^(-11) m
Now, let's calculate the change in wavelength using the Compton scattering formula:
Δλ = λ_c * (1 - cos(θ))
Substituting the values:
Δλ = (2.43 × 10^(-12) m) * (1 - cos(57.7 degrees))
Calculating cos(57.7 degrees):
cos(57.7 degrees) ≈ 0.551
Δλ = (2.43 × 10^(-12) m) * (1 - 0.551) ≈ 1.09 × 10^(-12) m
Finally, we can calculate the wavelength of the scattered photon by subtracting the change in wavelength from the wavelength of the incident X-ray photon:
λ' = λ - Δλ
Substituting the values:
λ' = (1.22 × 10^(-11) m) - (1.09 × 10^(-12) m) ≈ 1.11 × 10^(-11) m
Therefore, the wavelength of the scattered photon is approximately 1.11 × 10^(-11) meters.
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This table shows Wayne’s weight on four different planets.
Planet Wayne’s weight
(pounds)
Mars 53
Neptune 159
Venus 128
Jupiter 333
Arrange the planets in decreasing order of their strength of gravity.
Answer: Jupiter > Neptune > Venus > Mars
Explanation: edmentum
The deep end of a pool is 2.67 meters. What is the water pressure at the bottom of the deep end? Density of water: 1000 kg/m3
The water pressure at the bottom of the deep end of the pool is 26,370 Pascals (Pa).
To calculate the water pressure, we can use the formula:
Pressure = Density × Gravity × Height
Density of water = 1000 kg/m^3
Height = 2.67 meters
Gravity = 9.8 m/s^2 (approximate value)
Plugging in the values:
Pressure = 1000 kg/m^3 × 9.8 m/s^2 × 2.67 meters
Pressure ≈ 26,370 Pa
Therefore, the water pressure at the bottom of the deep end of the pool is approximately 26,370 Pascals.
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The velocity field of a flow is given by v= 6xi+ 6yj-7 tk.
a) Determine the velocity at a point x= 10 m; y = 6m; when t = 10 sec. Draw, approximately, a set of streamlines for the flow at instant t = 0.
b) Determine the acceleration field of the flow and the acceleration of the particle at the point and instant specified above. at the point and instant specified above
" The velocity at the point (x = 10 m, y = 6 m, t = 10 s) is 60i + 36j - 70k m/s.The acceleration of the particle at the point (x = 10 m, y = 6 m, t = 10 s) is -7k m/s²." Acceleration is a fundamental concept in physics that measures the rate of change of velocity of an object over time. It is defined as the derivative of velocity with respect to time.
a) To determine the velocity at the specified point (x = 10 m, y = 6 m, t = 10 s), we substitute these values into the given velocity field equation:
v = 6xi + 6yj - 7tk
v = 6(10)i + 6(6)j - 7(10)k
= 60i + 36j - 70k
Therefore, the velocity at the point (x = 10 m, y = 6 m, t = 10 s) is 60i + 36j - 70k m/s.
b) The acceleration field (a) can be obtained by taking the time derivative of the velocity field:
a = dv/dt = d(6xi + 6yj - 7tk)/dt
= 6(dxi/dt) + 6(dyj/dt) - 7(dtk/dt)
= 6(0i) + 6(0j) - 7k
= -7k
Therefore, the acceleration field is a = -7k m/s².
To determine the acceleration of the particle at the specified point (x = 10 m, y = 6 m, t = 10 s), we substitute these values into the acceleration field equation:
a = -7k
a = -7(1)k
= -7k
So, the acceleration of the particle at the point (x = 10 m, y = 6 m, t = 10 s) is -7k m/s².
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6. A mass density p = p(x, t) obeys the physical law j = vop where > 0 is a constant and j is the mass density flux. Use the continuity law, in the absence of any source or sink terms, to obtain a differential equation for p. The system is initially primed such that p(x,0) = poe-²/ where po, l are (positive) constants. Use the method of characteristics to determine the mass density for times t > 0. Sketch the profile of p against æ for a variety of time steps. [15 marks] Describe the significance of each of the quantities vo. Po and l. Illustrate each with a sketch at an appropriate number of time steps. [5 marks]
The continuity law and the physical law j = vop, we can derive a differential equation for the mass density p(x, t). The significance of the quantities vo, po, and l are that vo represents the velocity of the characteristic curves, po is the initial mass density at t = 0 and l is a positive constant.
The system is initially primed with a given initial condition p(x, 0) = po * e^(-x^2), where po and l are positive constants. The method of characteristics can be applied to determine the mass density for times t > 0 and sketch its profile against x for different time steps. The quantities vo, po, and l have specific meanings and significance in the context of the problem.
The continuity law states that the rate of change of mass density p with respect to time t plus the divergence of the mass density flux j must be zero in the absence of any source or sink terms.
Applying this law to the physical law j = vop, where v and o are constants, we have:
∂p/∂t + ∂(vop)/∂x = 0
Expanding the equation, we get:
∂p/∂t + vo ∂p/∂x + vop ∂o/∂x = 0
Since the system is initially primed with p(x, 0) = po * e^(-x^2), we have an initial condition for the mass density.
To solve this differential equation for times t > 0, we can use the method of characteristics. This method involves defining characteristic curves that satisfy the equation:
dx/dt = vo
By solving this equation, we can determine the characteristics curves and track the behavior of the mass density along these curves.
The significance of the quantities vo, po, and l can be described as follows:
- vo represents the velocity of the characteristic curves. It determines the speed at which the mass density propagates along these curves.
- po is the initial mass density at t = 0. It represents the value of the mass density at the initial condition.
- l is a positive constant that likely represents a characteristic length scale in the system.
By sketching the profile of p against x for different time steps, we can observe how the mass density evolves and propagates in space over time, following the characteristics curves determined by the initial conditions and the physical laws governing the system.
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A source emits sound waves in all directions.
The intensity of the waves 4.00 m from the sources is 9.00 *10-4 W/m?
Threshold of Hearing is 1.00 * 10-12 W/m?
A.) What is the Intensity in decibels?
B.) What is the intensity at 10.0 m from the source in Watts/m2?
C.) What is the power of the source in Watts?
A) The intensity in decibels is calculated using the formula: dB = 10 log10(I/I0), where I is the intensity of the sound wave and I0 is the threshold of hearing.
B) To find the intensity at 10.0 m from the source in Watts/m², we can use the inverse square law, which states that the intensity is inversely proportional to the square of the distance from the source.
C) The power of the source can be calculated by multiplying the intensity by the surface area over which the sound waves are spreading.
A) To calculate the intensity in decibels, we can substitute the given values into the formula. Using I = 9.00 * 10⁽⁻⁴⁾ W/m² and I0 = 1.00 * 10⁽⁻¹²⁾ W/m², we can find dB = 10 log10(9.00 * 10⁽⁻⁴⁾ / 1.00 * 10⁽⁻¹²⁾).
B) Applying the inverse square law, we can determine the intensity at 10.0 m from the source by multiplying the initial intensity (9.00 * 10⁽⁻⁴⁾ W/m²) by (4.00 m)² / (10.0 m)².
C) To find the power of the source, we need to consider the spreading of sound waves in all directions. Since the intensity at a distance of 4.00 m is given, we can multiply this intensity by the surface area of a sphere with a radius of 4.00 m.
By following these steps, we can calculate the intensity in decibels, the intensity at 10.0 m from the source, and the power of the source in Watts.
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A block is sliding with constant acceleration down. an incline. The block starts from rest at f= 0 and has speed 3.40 m/s after it has traveled a distance 8.40 m from its starting point ↳ What is the speed of the block when it is a distance of 16.8 m from its t=0 starting point? Express your answer with the appropriate units. μA 3 20 ? 168 Value Units Submit Request Answer Part B How long does it take the block to slide 16.8 m from its starting point? Express your answer with the appropriate units.
Part A: The speed of the block when it is a distance of 16.8 m from its starting point is 6.80 m/s. Part B: The time it takes for the block to slide 16.8 m from its starting point is 2.47 seconds.
To find the speed of the block when it is a distance of 16.8 m from its starting point, we can use the equations of motion. Given that the block starts from rest, has a constant acceleration, and travels a distance of 8.40 m, we can find the acceleration using the equation v^2 = u^2 + 2as. Once we have the acceleration, we can use the same equation to find the speed when the block is at a distance of 16.8 m. For part B, to find the time it takes to slide 16.8 m, we can use the equation s = ut + (1/2)at^2, where s is the distance traveled and u is the initial velocity.
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