A mechanic pushes a 2.10×10^ 3 −kg car from rest to a speed of v, doing 5,040 J of work in the process. During this time, the car moves 27.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car. (a) the speed v m/s (b) the horizontal force exerted on the car (Enter the magnitude.)

In order to answer the questions, we need more information about the inductor, such as its inductance value and any resistance in the circuit. The time constant and current can be determined using the formula for an RL circuit, which is given by:

I(t) = (V/R) * (1 - e^(-t/τ))

Where:

I(t) is the current at time t,

V is the voltage across the inductor,

R is the resistance in the circuit,

τ is the time constant, and

e is the base of the natural logarithm.

Part (a) - Time Constant:

To calculate the time constant of the inductor, we need to know the inductance (L) and resistance (R) in the circuit. The time constant (τ) is given by the formula:

τ = L / R

Once we have the values of L and R, we can calculate the time constant.

Part (b) - Current after 13 ms:

Using the formula mentioned earlier, we can substitute the values of V (12.0 V), R, and τ into the equation to calculate the current (I) at t = 13 ms.

Without the values for inductance and resistance, we cannot provide specific answers. Please provide the missing values so that we can assist you further in calculating the time constant and current in the circuit.

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The plane needs to take a bearing of 235.19 degrees to reach its destination.

Northward component = 25 km/h * sin(215 degrees) ≈ -16.45 km/h

Eastward component = 25 km/h * cos(215 degrees) ≈ -14.87 km/h

Northward component = 5 km/h * sin(210 degrees) ≈ -2.58 km/h

Eastward component = 5 km/h * cos(210 degrees) ≈ -4.33 km/h (opposite

Total northward component = -16.45 km/h + (-2.58 km/h) ≈ -19.03 km/h

Total eastward component = -14.87 km/h + (-4.33 km/h) ≈ -19.20 km/h

Resultant ground speed = sqrt((-19.03 km/h)^2 + (-19.20 km/h)²) ≈ 26.93 km/h

Resultant direction = atan((-19.20 km/h) / (-19.03 km/h)) ≈ 135.19 degrees

Final bearing = 135.19 degrees + 100 degrees

≈ 235.19 degrees

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When electrons vibrate sympathetically in a radio wave, this is an example of resonance.

Resonance is a particular form of mechanical wave motion that occurs when an external force is added to a system at its natural frequency, causing it to oscillate at a higher amplitude. The amplitude of the vibration grows exponentially until a maximum value is reached when resonance occurs.

When electrons vibrate sympathetically in a radio wave, this is an example of resonance. In general, resonances occur when the frequency of a driving force is the same as that of a natural frequency of a system. When a system is exposed to a periodic stimulus, the system will oscillate with an amplitude that is proportional to the strength of the stimulus at its natural frequency.

The passage above explains what resonance is and what happens when a system oscillates at a higher amplitude. Therefore, the best answer to the given question is "C. resonance."

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Each capacitor will store the same amount of energy which is 72 J.

Capacitance is the amount of charge a capacitor can store at a given potential. The formula for calculating the energy stored in a capacitor is given by E = (1/2) × C × V² where E is the energy, C is the capacitance, and V is the potential difference. In the given problem, two identical 1.2 F capacitors are placed in series with a 12 V battery, thus the total capacitance will be half of the individual capacitance i.e. 0.6 F. Using the formula above, we get

E = (1/2) × 0.6 F × (12 V)²= 43.2 J.

This is the total energy stored in both capacitors. Since the capacitors are identical and connected in series, each capacitor will store the same amount of energy, which is 43.2 J ÷ 2 = 21.6 J. Therefore, the energy stored in each capacitor is 21.6 J.

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Near point = 14.0 cm Far point = 119 cm Distance between retina and eye lens = 2.00 cm

The distance between the near point and the eye lens is = 14 - 2 = 12 cm

The distance between the far point and the eye lens is = 119 - 2 = 117 cm

Lens formula,1/f = 1/v - 1/u Where,f = focal length of the eye lens v = distance of far point u = distance of near point

Therefore, 1/f = 1/119 - 1/14= (14 - 119) / 14 × 119= - 105 / 1666f = - 1666 / (-105) = 15.876 cm

Therefore, The focal length of the eye lens is = 15.876 cm

Now, The power of the eye lens, P = 1/f= 1/15.876= 0.063 diopters

The formula for lens power is, P = 1/f or f = 1/P

Therefore, f = 1/0.063= 15.876 cm

Here, The power of the eyeglass lens the child should use for relaxed distance vision is = - 2.34 diopters.

Now, The image formed by the eye lens is a real and inverted image, which means that the eye lens is a converging lens.

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The frequency received by the observers is 55.78 Hz. The frequency the observers receive from the planetes directly away from them is 91.43 Hz.

(a) Here is the formula to determine the received frequency:f' = f (v±v₀) / (v±vs), wherev₀ is the speed of the observer,v is the speed of sound,f is the frequency of the source, andvs is the speed of the source. Here is the solution to part (a): The speed of sound is given as 342 m/s. Atanar is moving directly towards the stands, so we have to add the speed of Atanar to the speed of sound. The speed of Atanar is 1130 km/h, which is 313.8889 m/s when converted to m/s.v = 342 m/s + 313.8889 m/s = 655.8889 m/sUsing the formula,f' = f (v±v₀) / (v±vs),we get:f' = 60 Hz (655.8889 m/s) / (655.8889 m/s + 0 m/s)f' = 55.78 HzSo, the frequency received by the observers is 55.78 Hz.

(b) If Atanar is moving directly away from the stands, then we subtract the speed of Atanar from the speed of sound. Using the formula:f' = f (v±v₀) / (v±vs),we get:f' = 60 Hz (655.8889 m/s) / (655.8889 m/s - 0 m/s)f' = 91.43 Hz.Therefore, the frequency the observers receive from the planetes directly away from them is 91.43 Hz.

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The force exerted by the water on the bottom of tank A is less than the force exerted by the water on the bottom of tank B.

The force exerted by a fluid depends on its pressure and the surface area it acts upon. In this case, although the water level and height of the tanks are equal, tank A has the greatest surface area at the bottom, tank B has a middle surface area, and tank C has the least surface area.

The force exerted by the water on the bottom of a tank is directly proportional to the pressure and the surface area. Since the water pressure at the bottom of the tanks is the same (as they are filled to the same depth), the force exerted by the water on the bottom of tank A would be greater than the force exerted on tank B because tank A has a larger surface area at the bottom.

The pressure exerted on the bottom of tank A is equal to the pressure on the bottom of the other two tanks. Pressure in a fluid is determined by the depth of the fluid and the density of the fluid, but it is not affected by the surface area. Therefore, the pressure at the bottom of all three tanks is the same, regardless of their surface areas.

The force due to the water on the bottom of tank B is true and equal to the weight of the water in the tank. This is because the force exerted by a fluid on a surface is equal to the weight of the fluid directly above it. In tank B, the water exerts a force on its bottom that is equal to the weight of the water in the tank.

The water in tank C does not exert a downward force on the sides of the tank. The pressure exerted by the water at any given depth is perpendicular to the sides of the container. The force exerted by the water on the sides of the tank is a result of the pressure, but it acts horizontally and is balanced out by the pressure from the opposite side. Therefore, the water in tank C exerts an equal pressure on the sides of the tank but does not exert a net downward force.

The pressure at the bottom of tank A is less than the pressure at the bottom of tank C. This is because pressure in a fluid increases with depth. Since tank A has a greater depth than tank C (as they are filled to the same level), the pressure at the bottom of tank A is greater.

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The inductance of the solenoid is approximately 5.02 × 10^-4 Henrys (H).

The inductance of a solenoid can be calculated using the formula:

L = (μ₀ * N² * A) / l

where L is the inductance, μ₀ is the permeability of free space (4π × 10^-7 T·m/A), N is the number of turns, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.

Given:

N = 42 turns

r = 1.8 mm = 1.8 × 10^-3 m (radius)

l = 1.31 cm = 1.31 × 10^-2 m (length)

The cross-sectional area A of the solenoid can be calculated as:

A = π * r²

Substituting the values into the formula:

A = π * (1.8 × 10^-3 m)²

A ≈ 3.23 × 10^-6 m²

Now, we can calculate the inductance L:

L = (4π × 10^-7 T·m/A) * (42 turns)² * (3.23 × 10^-6 m²) / (1.31 × 10^-2 m)

L ≈ 5.02 × 10^-4 H

Therefore, the inductance of the solenoid is approximately 5.02 × 10^-4 Henrys (H).

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Maximum vertical component of initial velocity (vy0) at t = 0: 19.6 m/s. and Horizontal component of initial velocity (vx0) at t = 0: 122.5 m/s.

To calculate the maximum vertical component of the initial velocity (vy0) at t = 0 needed to touch the top of the building, we can use the equation of motion for vertical motion. The projectile needs to reach a height of 325 meters, so the maximum vertical displacement (Δy) is 325 meters. Since we're ignoring air resistance, the only force acting vertically is gravity. Using the equation Δy = vy0 * t + (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2), we can rearrange the equation to solve for vy0. At the maximum height, the vertical displacement is zero, so the equation becomes 0 = vy0 * t - (1/2) * g * t^2. Substituting the values, we have 0 = vy0 * t - (1/2) * 9.8 * t^2. Solving this quadratic equation, we find t = 2s (taking the positive root). Plugging this value into the equation, we can solve for vy0: 0 = vy0 * 2s - (1/2) * 9.8 * (2s)^2. Solving for vy0, we get vy0 = 9.8 * 2s = 19.6 m/s. (b) To calculate the horizontal component of the initial velocity (vx0) at t = 0 needed for the projectile to move 245 m and touch the top of the building, we can use the equation of motion for horizontal motion. The horizontal distance (Δx) the projectile needs to travel is 245 meters. The horizontal component of the initial velocity (vx0) remains constant throughout the motion since there are no horizontal forces acting on the projectile. Using the equation Δx = vx0 * t, we can rearrange the equation to solve for vx0. Since the time of flight is the same for both the vertical and horizontal motions (2s), we can substitute the value of t = 2s into the equation. Thus, we have 245 = vx0 * 2s. Solving for vx0, we get vx0 = 245 / (2s) = 122.5 m/s.

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The electric field flux through the square is determined as 2.25 x 10⁵ Nm²/C.

The electric field flux through the square is calculated by applying the following formula as follows;

Ф = EA

where;

The magnitude of the electric field is calculated as;

E = (kQ) / s²

E = ( 9 x 10⁹ x 25 x 10⁻⁶ ) / ( 0.15 m)²

E = 1 x 10⁷ N/C

The electric field flux through the square is calculated as;

Ф = EA

Ф = (1 x 10⁷ N/C) x (0.15 m)²

Ф = 2.25 x 10⁵ Nm²/C

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The magnitude of the positive charge is 4.58×10−7 C and the magnitude of the negative charge is 2.97×10−7 C.

Let's denote the magnitude of the positive charge as q1 and the magnitude of the negative charge as q2. Then, we can apply Coulomb's law to the initial situation where the spheres are separated by 0.4 m and attracting each other with a force of 0.244 N:

[tex]$$F = k\frac{q_1q_2}{r^2}$$$$0.244 = k\frac{q_1q_2}{0.4^2}$$[/tex]

where k is the Coulomb constant. We don't need to know the value of k, we just need to know that it's a constant.

We can simplify the equation above and express q2 in terms of q1:

[tex]$$F = k\frac{q_1q_2}{r^2}$$$$0.244 = k\frac{q_1q_2}{0.4^2}$$[/tex]

Now, when the spheres are connected by a thin conducting wire and then removed, they will have the same potential. Therefore, they will share the charge equally. The final force between them is 0.035 N and is repulsive.

We can apply Coulomb's law again:

[tex]$$F = k\frac{q^2}{r^2}$$$$0.035 = k\frac{(q_1+q_2)^2}{0.4^2}$$[/tex]

where q is the charge on each sphere. We can substitute the expression for q2 that we found earlier:

[tex]$$0.035 = k\frac{(q_1+\frac{0.244\cdot0.4^2}{kq_1})^2}{0.4^2}$$[/tex]

This is a quadratic equation in q1. We can solve it to find

[tex]q1:$$q_1 = 4.58\times10^{-7} \ C$$[/tex]

Thus, the magnitude of the positive charge is 4.58×10−7 C and the magnitude of the negative charge is 2.97×10−7 C.

When they are separated by a distance of 0.4 m, they attract each other with a force of 0.244 N.

Coulomb's law can be applied in this initial situation.

[tex]$$F = k\frac{q_1q_2}{r^2}$$$$0.244 = k\frac{q_1q_2}{0.4^2}$$[/tex]

Here, k is the Coulomb constant. The magnitude of the positive charge can be denoted as q1 and that of the negative charge as q2. The expression for q2 in terms of q1 can be derived from the equation above. We obtain:

[tex]$$q_2 = \frac{0.244\cdot0.4^2}{kq_1}$$[/tex]

Now, the spheres are connected by a thin conducting wire, and they will share the charge equally.

Therefore, the final force between them is repulsive and 0.035 N. Again, Coulomb's law can be applied:

[tex]$$F = k\frac{q^2}{r^2}$$$$0.035 = k\frac{(q_1+q_2)^2}{0.4^2}$$[/tex]

[tex]$$0.035 = k\frac{(q_1+\frac{0.244\cdot0.4^2}{kq_1})^2}{0.4^2}$$[/tex]

This is a quadratic equation in q1, which can be solved to find that the magnitude of the positive charge is 4.58×10−7 C, and that of the negative charge is 2.97×10−7 C.

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The magnitude of the net electric force on each particle is 2.025 N directed away from the triangle.

Charge on each particle, q1 = q2 = q3 = -15 × 10⁻⁶C

∴ Net force on particle 1 = F1

Net force on particle 2 = F2

Net force on particle 3 = F3

The magnitude of the net electric force on each particle:

It can be determined by using Coulomb's Law:

F = kqq / r²

where

k = Coulomb's constant = 9 × 10⁹ Nm²/C²

q = charge on each particle

r = distance between the particles

We know that all three charges are negative, so they will repel each other. Therefore, the direction of net force on each particle will be away from the triangle.

From the given data,

Side of equilateral triangle, a = 25cm = 0.25m

∴ Distance between each corner of the triangle = r = a = 0.25m

∴ Net force on particle 1 = F1

F1 = kq² / r² = 9 × 10⁹ × (-15 × 10⁻⁶)² / (0.25)²= -2.025 N

∴ Net force on particle 2 = F2

F2 = kq² / r² = 9 × 10⁹ × (-15 × 10⁻⁶)² / (0.25)²= -2.025 N

∴ Net force on particle 3 = F3

F3 = kq² / r² = 9 × 10⁹ × (-15 × 10⁻⁶)² / (0.25)²= -2.025 N

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The electric field's E magnitude at the midpoint between the two disks is 3.6 x 10⁷ N/C.

When two charged plates face each other, they form a capacitor. The electric field at the midpoint of two plates is provided by the expression for a parallel plate capacitor:

Electric field, E = σ/2εwhere σ is the surface charge density, and ε is the permittivity of the space or material between the plates.In this question, both plates are circular with a diameter of 10cm.

So, we can calculate the surface area of each plate by using the equation for the area of a circle:

A = πr²

where r is the radius of the circle, given as 5cm.

A = π(5cm)² = 78.5cm²

The surface charge density is given in nano-coulombs (nC), so we need to convert it to Coulombs (C).

1nC = 1 x 10⁻⁹C

Because the left plate is charged to -50nC, the surface charge density is:-

50nC / 78.5cm² = -6.37 x 10⁻¹⁰C/cm²

Because the right plate is charged to +50nC, the surface charge density is:

+50nC / 78.5cm² = 6.37 x 10⁻¹⁰C/cm²

The electric field at the midpoint between the two plates can now be calculated:

|E| = σ/2ε = 6.37 x 10⁻¹⁰C/cm² / (2 x 8.85 x 10⁻¹²F/cm) = 3.6 x 10⁷N/C

Due to the nature of the problem, the electric field between the two plates is directed from right to left, and its magnitude is 3.6 x 10⁷ N/C (newtons per coulomb).

Therefore, the magnitude of the electric field at the midpoint between the two disks is 3.6 x 10⁷ N/C.

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a. The work done by the applied force is approximately 1380.3 Joules.

b. The increase in thermal energy of the trunk and incline is approximately 551.2 Joules.

a. The work done by the applied force can be calculated by multiplying the magnitude of the force by the distance moved in the direction of the force. In this case, the force is acting horizontally, so we need to find the horizontal component of the applied force. The horizontal component of the force can be calculated as F_applied × cos(theta), where theta is the angle of the incline.

F_applied = m × g × sin(theta),

F_horizontal = F_applied × cos(theta).

Plugging in the values:

m = 50 kg,

g = 9.8 m/s² (acceleration due to gravity),

theta = 31 degrees.

F_applied = 50 kg × 9.8 m/s² × sin(31 degrees) ≈ 246.2 N.

F_horizontal = 246.2 N × cos(31 degrees) ≈ 212.2 N.

The work done by the applied force is given by:

Work = F_horizontal × distance,

Work = 212.2 N × 6.5 m ≈ 1380.3 Joules.

Therefore, the work done by the applied force is approximately 1380.3 Joules.

b. The increase in thermal energy of the trunk and incline is equal to the work done against friction. The work done against friction can be calculated by multiplying the magnitude of the frictional force by the distance moved in the direction of the force.

Frictional force = coefficient of kinetic friction × normal force,

Normal force = m × g × cos(theta).

Plugging in the values:

Coefficient of kinetic friction = 0.20,

m = 50 kg,

g = 9.8 m/s² (acceleration due to gravity),

theta = 31 degrees.

Normal force = 50 kg × 9.8 m/s² × cos(31 degrees) ≈ 423.9 N.

Frictional force = 0.20 × 423.9 N ≈ 84.8 N.

The increase in thermal energy is given by:

Thermal energy = Frictional force × distance,

Thermal energy = 84.8 N × 6.5 m ≈ 551.2 Joules.

Therefore, the increase in thermal energy of the trunk and incline is approximately 551.2 Joules.

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The maximum torque that the coil can experience can be calculated using the formula:

τ = N * B * A * sin(θ)

where τ is the torque, N is the number of turns, B is the magnetic field, A is the area of the coil, and θ is the angle between the magnetic field and the normal to the coil.

In this case, the coil has one turn (N = 1), a magnetic field of 2.56 T, and the length of the wire is used to make a circular coil, so the perimeter of the coil is equal to the length of the wire.

The perimeter of the coil (P) is given by:

P = 2πr

where r is the radius of the coil.

Since there is one turn, the circumference of the coil is equal to the length of the wire:

P = L

where L is the length of the wire.

Therefore, we can find the radius of the coil (r) using the formula:

r = L / (2π)

Substituting the given values:

r = (7.99 x 10^-2 m) / (2π)

Now we can calculate the area of the coil (A):

A = πr^2

Substituting the value of r:

A = π * [(7.99 x 10^-2 m) / (2π)]^2

Finally, we can calculate the maximum torque:

τ = (1) * (2.56 T) * A * sin(θ)

Since the problem does not specify the angle θ, we assume it to be 90 degrees to maximize the torque:

τ = (2.56 T) * A

Substituting the value of A:

τ = (2.56 T) * [π * [(7.99 x 10^-2 m) / (2π)]^2]

τ ≈ 5.22 x 10^-3 N·m

Therefore, the maximum torque that this coil can experience is approximately 5.22 x 10^-3 N·m.

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The surface contamination level is determined to be 540 counts, taking into account the uncorrected count rate on the smear paper, background count rate, counting system efficiency, area of the smeared surface, and pick-up efficiency of the smear.

To calculate the surface contamination level, we need to consider the count rate on the smear paper, the background count rate, the efficiency of the counting system, the area of the surface smeared, and the pick-up efficiency of the smear.

Given:

Uncorrected count rate on smear paper = 3840 counts/min

Background count rate = 240 counts/min

Efficiency of counting system = 15%

Area of surface smeared = 0.1 m²

Pick-up efficiency of smear = 10%

First, we need to correct the count rate on the smear paper by subtracting the background count rate:

Corrected count rate = Uncorrected count rate - Background count rate

Corrected count rate = 3840 counts/min - 240 counts/min

Corrected count rate = 3600 counts/min

Next, we need to calculate the total number of counts on the surface:

Total counts = Corrected count rate * Efficiency of counting system * Area of surface smeared

Total counts = 3600 counts/min * 0.15 * 0.1 m²

Total counts = 54 counts

Finally, we can calculate the surface contamination level:

Contamination level = Total counts * (1 / Pick-up efficiency of smear)

Contamination level = 54 counts * (1 / 0.10)

Contamination level = 540 counts

Therefore, the surface contamination level is 540 counts.

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The cannonball, fired from ground level with an initial speed of 105 m/s at an angle of 26 degrees above the horizontal, will hit the ground at a certain distance of 276 meters.

To determine this distance, we can calculate the projectile's horizontal range using the given information.

The horizontal range of a projectile can be determined using the equation:

Range = (initial velocity^2 * sin(2 * launch angle)) / gravitational acceleration

In this case, the initial velocity is 105 m/s and the launch angle is 26 degrees. The gravitational acceleration is approximately 10 m/s^2. Plugging these values into the equation, we can calculate the range:

Range = (105^2 * sin(2 * 26)) / 10

Simplifying this expression, we get:

Range ≈ 276 meters

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The question asks for the current required to generate a magnetic field of 9.0e-5 T at the Earth's poles during a pole reversal. The current is generated by a loop around the equator, and we need to determine the magnitude of the current. The Earth's magnetic field currently has a magnitude of approximately 7 × 10-5 T at the poles.

To determine the current required to generate a magnetic field of 9.0e-5 T at the Earth's poles, we can use Ampere's law. Ampere's law relates the magnetic field generated by a current-carrying loop to the current and the distance from the loop. In this case, we want to generate a magnetic field at the poles, which are located at the ends of the Earth's diameter. The diameter of the Earth is given as 2 * RE, where RE is the radius of the Earth.

Since the current loop is placed around the equator, the distance from the loop to the poles is half the Earth's diameter, or RE. Therefore, we can use Ampere's law to solve for the current: B = (μ₀ * I) / (2 * π * R), where B is the desired magnetic field, μ₀ is the permeability of free space, I is the current, and R is the distance from the loop. Rearranging the equation to solve for I, we have: I = (B * 2 * π * R) / μ₀.

Substituting the given values, where B is 9.0e-5 T and R is 6.37 × 10^6 m, we can calculate the current required. Using the value for μ₀, which is approximately 4π × 10^-7 T·m/A, we can solve for I.

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The initial speed of the hammer thrown by Thor is approximately 105.82 m/s. To determine the initial speed and direction of the hammer thrown by Thor, we can use the principle of conservation of momentum and the concept of vector addition.

Let's denote the initial speed of the hammer as v₁ and its direction as θ₁. We'll assume the positive x-axis is to the right and the positive y-axis is upward.

According to the conservation of momentum:

(m₁ * v₁) + (m₂ * v₂) = (m₁ * u₁) + (m₂ * u₂)

where m₁ and m₂ are the masses of the hammer and Superman, v₁ and v₂ are their initial velocities, and u₁ and u₂ are their final velocities.

m₁ (mass of hammer) = 20 kg

v₂ (initial velocity of Superman) = -20 m/s (negative sign indicates downward direction)

m₂ (mass of Superman) = 100 kg

u₁ (final velocity of hammer) = 10 m/s (speed)

u₂ (final velocity of Superman) = 10 m/s (speed)

θ₂ (angle of motion of Superman) = 40 degrees above the horizontal

Now, let's calculate the initial velocity of the hammer.

Using the conservation of momentum equation and substituting the given values:

(20 kg * v₁) + (100 kg * (-20 m/s)) = (20 kg * 10 m/s * cos(θ₂)) + (100 kg * 10 m/s * cos(40°))

Note: The negative sign is applied to the velocity of Superman (v₂) since it is directed downward.

Simplifying the equation:

20 kg * v₁ - 2000 kg m/s = 200 kg * 10 m/s * cos(θ₂) + 1000 kg * 10 m/s * cos(40°)

Now, solving for v₁:

20 kg * v₁ = 2000 kg m/s + 200 kg * 10 m/s * cos(θ₂) + 1000 kg * 10 m/s * cos(40°)

v₁ = (2000 kg m/s + 200 kg * 10 m/s * cos(θ₂) + 1000 kg * 10 m/s * cos(40°)) / 20 kg

Calculating the value of v₁:

v₁ ≈ 105.82 m/s

Therefore, the initial speed of the hammer thrown by Thor is approximately 105.82 m/s.

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Since the crate is sliding down due to gravity, the force parallel to the incline acting on the crate is less than the maximum static frictional force acting on it

In order for the crate to slide with an acceleration of 3.85 m/s²,

The angle of the incline must be 20.7°.

Explanation: Given data;

Mass of the crate, m = 35.0 kg

Coefficient of kinetic friction, μ = 0.268

Acceleration, a = 3.85 m/s²

The forces acting on the crate are; The force due to gravity, Fg = mg

The force acting on the crate parallel to the incline, F∥The force acting perpendicular to the incline, F⊥The normal force acting on the crate is equal to and opposite to the perpendicular force acting on it.

Therefore;F⊥ = mgThe force acting parallel to the incline is;F∥ = ma

Since the crate is sliding down due to gravity, the force parallel to the incline acting on the crate is less than the maximum static frictional force acting on it. The maximum force of static friction, f max, is given by fmax = N, where N is the normal force acting on the crate.

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The rms current in the circuit is approximately 0.189 A.

The question requires us to calculate the rms current of a circuit that consists of a resistor, an inductor, and a capacitor in series. The circuit is driven by an AC voltage source that has a frequency of 876 cycles/s and an amplitude of 190 V.Let's begin by finding the total impedance of the circuit. The impedance of a series RLC circuit is given by:Z = R + j(XL - XC)where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. The imaginary part of the impedance represents the reactance of the circuit, which depends on the frequency of the applied voltage. At resonance, XL = XC, and the total impedance is equal to the resistance Z = R.

To calculate the impedance of the circuit, we need to find the values of XL and XC at the given frequency f = 876 cycles/s. The inductive reactance is given by:XL = 2πfLwhere L is the inductance of the inductor. Substituting the given values, we get:XL = 2π(876)(72 × 10⁻³) = 101.94 ΩThe capacitive reactance is given by:XC = 1/(2πfC)where C is the capacitance of the capacitor. Substituting the given values, we get:XC = 1/(2π(876)(0.3 × 10⁻⁶)) = 607.71 ΩThe total impedance is therefore:Z = R + j(XL - XC) = 108 + j(-505.77) = 108 - j505.77.

The rms current is given by the ratio of the rms voltage to the impedance:Irms = Vrms/Zwhere Vrms is the rms value of the applied voltage. The rms value of a sinusoidal voltage is given by the peak voltage divided by the square root of 2 (Vrms = Vpeak/√2). Substituting the given values, we get:Vrms = 190/√2 = 134.35 VIrms = Vrms/Z = 134.35/(108 - j505.77) = 0.189 - j0.886 ARms current, Irms = 0.189 A (approx).

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The frequency can be expressed as [tex]4.366 *10^{14} Hz[/tex]the wavelength λ  can be expressed as [tex]6.876 *10^{-7} meters[/tex]

The frequency of a repeated event is its number of instances per unit of time. For clarity and to distinguish it from spatial frequency, it is also sometimes referred to as temporal frequency.

Frequency is measured in hertz which is equal to one event per secondGiven that Energy =2.89×10 −19 J

h = plank constant = [tex]6.626 *10^{-34}[/tex]

E = hf

f = E / h

f = [tex]\\\frac{2.89* 10^{-19} }{ 6.626*10^{-34} }[/tex]

f= [tex]4.366 *10^{14} Hz[/tex]

To calculate the wavelength we can use

λ = c / f

λ = [tex]\\\frac{2.998 *10^8}{4.366*10^14}[/tex]

λ =[tex]6.876 *10^-7 meters[/tex]

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The force of gravity acting on the masses is given by the formula;

F = Gm₁ m₂/r²

where G is the gravitational constant, m₁ and m₂ are the masses, and r is the distance between the masses.

Since the electric force must be equal to the gravitational force,

F₁ = F₂ = Gm₁ m₂/r²

where F₁ is the electric force on one mass and F₂ is the electric force on the other mass.

Since the two masses are to have the same charge (q),

the electric force on each mass can be given by the formula.

F = kq²/r²

where k is the Coulomb constant, and q is the charge on each mass.

Similarly,

F₁ = F₂ = kq²/r²

Combining the two equations.

kq²/r² = Gm₁ m₂/r²

Dividing both sides by r².

kq²/m₁ m₂ = G

Now, the charges on the masses can be given by

q = √ (Gm₁ m₂/k)

Substituting the given values, and using the fact that the mass of each sphere is given by.

m = (4/3)πr³ρ

where ρ is the density, and r is the radius.

q = √ (6.67 × 10^-11 × 64 × 64 / 9 × 10^9)

q = √ 291.56q = 17.06 × 10^-9 C (to one decimal place)

the charge on each mass must be 17.06 nm.

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The initial speed of the ball i such that it will stop after t = 1 s is -9.8 m/s, and the space traveled by the ball when it stops is 8.48 m.

At t = 0, a ball is kicked such that it moves along a ramp that makes an angle θ = 30 degree with the ground.

Given that there is no friction between the ball and the ramp, we need to calculate the initial speed of the ball i such that it will stop after t = 1 s.

We also need to calculate the space traveled by the ball when it stops.

angle of the ramp θ = 30°

The horizontal component of the initial velocity of the ball is given as follows:

vₓ = vicosθvₓ = vi cosθ ………………….. (1)

The vertical component of the initial velocity of the ball is given as follows:

vᵧ = visinθ …………………………….. (2)

When the ball stops at t = 1 s,

its final velocity v = 0 m/s.

We know that the acceleration of the ball along the incline is given as follows:

a = gsinθ ………………………………..(3)

We also know that the time taken by the ball to stop is t = 1 s.

Therefore, we can find the initial velocity of the ball using the following formula:

v = u + at0 = vi + a*t

Substituting the values, we get:0 = vi + gsinθ*1

The initial velocity of the ball is given as follows:

vi = - gsinθ

The negative sign in the equation shows that the ball is decelerating.

The horizontal distance traveled by the ball is given as follows:

s = vₓ * t

The vertical distance traveled by the ball is given as follows:

h = vᵧ * t + 0.5*a*t²

We know that the ball stops at t = 1 s. Therefore, we can find the space traveled by the ball using the following formula:

s = vₓ * t

Substituting the values, we get:

s = vi cosθ * t

Therefore, the initial speed of the ball is given by:

vi = -g sinθ= -9.8 m/s

The space traveled by the ball when it stops is given by:

s = vₓ * t= vi cosθ * t= (-9.8 m/s) cos 30° × 1 s ≈ -8.48 m (since distance cannot be negative, the distance traveled by the ball is 8.48 m in the opposite direction).

Therefore, the initial speed of the ball i such that it will stop after t = 1 s is -9.8 m/s, and the space traveled by the ball when it stops is 8.48 m.

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The first order gives the best resolution. Thus, the correct answer is Option 2.

To determine the order that gives the best resolution for the given diffraction grating and wavelength, we can use the formula for the angular separation of the diffraction peaks:

θ = mλ / d,

where

θ is the angular separation,

m is the order of the diffraction peak,

λ is the wavelength of light, and

d is the spacing between the grating lines.

Given:

Wavelength (λ) = 623 nm

                         = 623 × 10⁻⁹ m,

Grating spacing (d) = 1 / (6900 lines/cm)

                               = 1 / (6900 × 10² lines/m)

                              = 1.449 × 10⁻⁵ m.

We can substitute these values into the formula to calculate the angular separation for different orders:

For zero order, θ₀ = (0 × 623 × 10^(-9) m) / (1.449 × 10^(-5) m),

                         θ₀ = 0

For first order θ₁ = (1 × 623 × 10^(-9) m) / (1.449 × 10^(-5) m),

                       θ₁  ≈ 0.0428 rad

For second-order θ₂ = (2 × 623 × 10^(-9) m) / (1.449 × 10^(-5) m)

                              θ₂  ≈ 0.0856 rad.

The angular separation determines the resolution of the diffraction pattern. Smaller angular separations indicate better resolution. Thus, the order that gives the best resolution is the order with the smallest angular separation. In this case, the best resolution is achieved in the first order,   θ₁  ≈ 0.0428 rad

Therefore, the correct answer is first order gives the best resolution.

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The proton's speed is approximately 1.48 x 10^5 m/s, which corresponds to option B) Counterclockwise.

We can use the formula for the centripetal force experienced by a charged particle moving in a magnetic field:

F = qvB

where F is the centripetal force, q is the charge of the particle, v is its velocity, and B is the magnetic field strength.

Since the proton moves in a circular path, the centripetal force is provided by the magnetic force:

F = mv^2/r

where m is the mass of the proton and r is the radius of the circular path.

Setting these two equations equal to each other, we have:

mv^2/r = qvB

Rearranging the equation, we find:

v = (qBr/m)^0.5

Plugging in the given values, we have:

v = [(1.6 x 10^-19 C)(9.8 x 10^-6 T)(4.95 x 10^-2 m)/(1.67 x 10^-27 kg)]^0.5

v ≈ 1.48 x 10^5 m/s

Therefore, the proton's speed is approximately 1.48 x 10^5 m/s.

Regarding the direction of the proton's motion as viewed from above, we can apply the right-hand rule. If the magnetic field is pointed into the page and the proton is moving to the left, the force experienced by the proton will be downwards. As a result, the proton will move in a counterclockwise direction, which corresponds to option B) Counterclockwise.

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The output gear is moving at 450 RPM. the speed of the driving gear is 112.5 RPM

To find the speed of the driving gear, we can use the concept of gear ratio. The gear ratio is defined as the ratio of the number of teeth on the driven gear to the number of teeth on the driving gear.

Given that the output gear has 72 teeth and there is a speed reduction of 4:1, we can calculate the number of teeth on the driving gear.

Number of teeth on the driving gear = Number of teeth on the driven gear / Speed reduction

Number of teeth on the driving gear = 72 teeth / 4 = 18 teeth

So, the driving gear has 18 teeth.

Now, to find the speed of the driving gear, we can use the formula:

Speed of the driving gear = Speed of the output gear / Speed reduction

Speed of the driving gear = 450 RPM / 4 = 112.5 RPM

Therefore, the speed of the driving gear is 112.5 RPM.

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Explanation:

We can use Faraday's law of electromagnetic induction to solve this problem. According to this law, the induced emf (ε) in a coil is equal to the negative of the rate of change of magnetic flux through the coil:

ε = - dΦ/dt

where Φ is the magnetic flux through the coil.

Rearranging this equation, we can solve for the magnetic flux:

dΦ = -ε dt

Integrating both sides of the equation, we get:

Φ = - ∫ ε dt

Since the emf and the rate of current change are constant, we can simplify the integral:

Φ = - ε ∫ dt

Φ = - ε t

Substituting the given values, we get:

ε = 15.0 mV = 0.0150 V

N = 513

di/dt = 10.0 A/s

i = 3.80 A

We want to find the magnetic flux through each turn of the coil at an instant when the current is 3.80 A. To do this, we first need to find the time interval during which the current changes from 0 A to 3.80 A:

Δi = i - 0 A = 3.80 A

Δt = Δi / (di/dt) = 3.80 A / 10.0 A/s = 0.380 s

Now we can use the equation for magnetic flux to find the flux through each turn of the coil:

Φ = - ε t = -(0.0150 V)(0.380 s) = -0.00570 V·s

The magnetic flux through each turn of the coil is equal to the total flux divided by the number of turns:

Φ/ N = (-0.00570 V·s) / 513

Taking the magnitude of the result, we get:

|Φ/ N| = 1.11 × 10^-5 V·s/turn

Therefore, the magnetic flux through each turn of the coil at the given instant is 1.11 × 10^-5 V·s/turn.

An ant stands 70 feet away from a tower, and has to look up at a 40 degree angle to see the top. The height of the tower is approximately 58.74 feet.

To find the height of the tower, we can use trigonometry. Let's denote the height of the tower as 'h'.

We have a right triangle formed by the ant, the tower, and the line of sight to the top of the tower. The distance from the ant to the base of the tower is 70 feet, and the angle formed between the ground and the line of sight is 40 degrees.

In a right triangle, the tangent function relates the opposite side to the adjacent side. In this case, the opposite side is the height of the tower (h), and the adjacent side is the distance from the ant to the tower (70 feet). Therefore, we can use the tangent function as follows:

tan(40°) = h / 70

To find the value of h, we can rearrange the equation:

h = 70 * tan(40°)

Now, let's calculate the height of the tower using the given formula:

h = 70 * tan(40°)

h ≈ 70 * 0.8391

h ≈ 58.7387 feet

Therefore, the height of the tower is approximately 58.74 feet.

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The wavelength of the of the harmonic wave traveling along the rope, given that it completes 38.0 vibrations in 32.0 s is 60.31 cm

First, we shall obtain the frequency of the wave. Details below:

Frequency (f) = Number of oscillation (n) / time (s)

= 38.0 / 32.0

= 1.18 Hertz

Next, we shall obtain the speed of the wave. Details below:

Speed = Distance / time

= 427 / 6

= 71.17 cm/s

Finally, we shall obtain the wavelength of the wave. Details below:

Speed (v) = wavelength (λ) × frequency (f)

71.17 = wavelength × 1.18

Divide both sides by 27×10⁸

Wavelength = 71.17 / 1.18

= 60.31 cm

Thus, the wavelength of the wave is 60.31 cm

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