a light string is wrapped around the rim of a small hoop if you hold the free end of the string in the hoop is released from rest it will unwind and the hoop descends, what force(s) is/are causing a torque on the hoop?

a-tension
b-weight
c-friction
d-normal force
e-more than one option is correct

Answers

Answer 1

Answer:

E: More than one option is correct  

Explanation:

Tension, weight, and friction produce torques on the hoop, while the normal force does not.


Related Questions

In a water pistol, a piston drives water through a larger tube of radius 1.30 cm into a smaller tube of radius 1.10 mm as in the figure below. Answer parts a-f.

Answers

It takes 0.47 seconds for water to travel from the nozzle to the ground when the water pistol is fired horizontally.

What is the time it takes for water to travel from the nozzle to the ground?

We will denote the height of the water pistol above the ground as h and the initial velocity of water exiting the nozzle as v2. Assuming negligible air resistance, we will analyze the vertical motion of the water droplets.

The vertical displacement of the water droplets is calculated using equation: h = (1/2) * g * t^2.

Rearranging equation, we solve for time:

t = sqrt(2h / g).

Given data:

Height h = 1.10 m and the acceleration due to gravity g = 9.8 m/s^2, we get:

t = sqrt(2 * 1.10 / 9.8)

t = 0.47380354147

t = 0.47 seconds.

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The temperature of a aluminum bar rises by 10.0°C when it absorbs 4.73 kJ of energy by heat. The mass of the bar is 525 g. Determine the specific heat of aluminum from these data. Answer is in kJ/kg · °C.

Answers

Answer:

Certainly! We can use the formula:

q = mcΔT

where q is the amount of heat absorbed, m is the mass of the aluminum bar, c is the specific heat capacity of aluminum, and ΔT is the change in temperature.

Substituting the given values, we get:

4.73 kJ = (0.525 kg) x c x (10.0°C)

Solving for c, we get:

c = 0.901 kJ/kg · °C

Therefore, the specific heat of aluminum is 0.901 kJ/kg · °C.

Explanation:

A satellite weighing 5,400 kg is launched into orbit 3.6400 x 107 m above the center of the earth.

The mass of Earth is 6.0 × 1024 kg. The gravitational constant is 6.673 × 10–11 N•m2/kg2.

The gravitational force of Earth on the satellite is ___



Group of answer choices

9.1 x 10^4

1.6 x 10^3

2.1 x 10^6

Answers

Answer:

[tex]\tt F=1.63*10^3 N[/tex]

Explanation:

Gravitational force is defined as the force of attraction between two objects with mass. It is a fundamental force of nature, and it is what keeps us on the ground and what keeps the planets in orbit around the Sun.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers

For the Question:

We can use the following formula to calculate the gravitational force between the Earth and the satellite:

[tex]\boxed{\tt F =\frac{ G * M * m }{ r^2}}[/tex]

Where:

F is the gravitational force

G is the gravitational constant[tex]\tt (6.673 * 10^{-11} Nm^2/kg^2)[/tex]

M is the mass of the Earth [tex]\tt (6.0 * 10^24 kg)[/tex]

m is the mass of the satellite[tex]\tt (5,400 kg)[/tex]

r is the distance between the satellite and the center of the Earth [tex]\tt (3.6400 * 10^7 m)[/tex]

Plugging in these values, we get the following:

[tex]\tt F = \frac{6.673 * 10^{-11} * 6.0 * 10^{24}* 5,400 }{ (3.6400 * 10^7 )^2}[/tex]

[tex]\tt F=1.63*10^3 N[/tex]

Therefore, answer is [tex]\tt F=1.63*10^3 N[/tex]

Joe is painting the floor of his basement using a paint roller. The roller has a mass of 2.4 kg and a radius of 3.8 cm. In rolling the roller across the floor, Joe applies a force F = 16 N directed at an angle of 35° as shown. Ignoring the mass of the roller handle, what is the magnitude of the angular acceleration of the roller?

Answers

The magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.

The magnitude of the angular acceleration of the roller can be determined using the torque equation and Newton's second law for rotational motion.
Step 1: Calculate the moment of inertia of the roller.
The moment of inertia (I) of a solid cylinder is given by the formula I = (1/2) * m * r^2, where m is the mass of the object and r is the radius.
In this case, the mass of the roller is 2.4 kg and the radius is 0.038 m.
So, I = (1/2) * 2.4 kg * (0.038 m)^2.
Step 2: Calculate the torque applied to the roller.
Torque (τ) is equal to the force (F) applied multiplied by the perpendicular distance (r) from the axis of rotation.
In this case, the force applied by Joe is 16 N and the distance is equal to the radius of the roller, 0.038 m.
So, τ = F * r.
Step 3: Use the torque equation.
The torque applied to the roller causes an angular acceleration (α) according to the equation τ = I * α.
Rearranging the equation, we get α = τ / I.
Step 4: Substitute the values into the equation.
Using the values we calculated earlier, we can substitute them into the equation α = τ / I.
α = (16 N * 0.038 m) / [(1/2) * 2.4 kg * (0.038 m)^2].
Step 5: Calculate the magnitude of the angular acceleration.
Evaluating the expression, we find that the magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.
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Someone goes to lift a crate that is resting on the bottom of the pool filled with water (density of water is 1000 kg/m^3). While
still submerged, only 310 N is required to lift the crate. The crate is shaped like a cube with sides of 0.25 m. What is the density of
the cube? Numerical answer is assumed to be in units of kg/m^3

Answers

Answer:

the density of the cube is approximately 2016.07 kg/m^3.

Explanation:

The buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

Let's first calculate the weight of the crate:

mass of crate = density * volume = density * (side length)^3 = density * 0.25^3 = 0.015625 * density

weight of crate = mass of crate * gravity = 0.015625 * density * 9.81 = 0.1530875 * density

where gravity is the acceleration due to gravity, which is approximately 9.81 m/s^2.

Since the crate is submerged in water, the buoyant force acting on it is:

buoyant force = weight of water displaced = density of water * volume of water displaced * gravity

The volume of water displaced is equal to the volume of the cube, which is 0.25^3 = 0.015625 m^3. Therefore, the buoyant force is:

buoyant force = 1000 kg/m^3 * 0.015625 m^3 * 9.81 m/s^2 = 1.534453125 N

According to the problem, it takes 310 N to lift the crate while it is still submerged. This means that the net force acting on the crate is:

net force = lifting force - buoyant force = 310 N - 1.534453125 N = 308.465546875 N

This net force is equal to the weight of the crate:

net force = weight of crate = 0.1530875 * density

Therefore, we can solve for the density of the crate:

density = net force / 0.1530875 = 308.465546875 / 0.1530875 = 2016.06666667 kg/m^3

Rounding to the nearest hundredth, we get:

density ≈ 2016.07 kg/m^3

Therefore, the density of the cube is approximately 2016.07 kg/m^3

. A 0.140 kg baseball is pitched toward home plate at 30.0 m/s.
The batter hits the ball back (opposite direction) to the pitcher at
44.0 m/s. Assume that towards home plate is positive. What is
the change in momentum for the ball?

Answers

The change in momentum for the baseball, which is hit back in the opposite direction by the batter, is -10.36 kg·m/s. This change in momentum is obtained by subtracting the initial momentum of 4.2 kg·m/s from the final momentum of -6.16 kg·m/s. The negative sign indicates the opposite direction of the momentum.

To find the change in momentum for the baseball, we can use the formula:

Change in momentum = Final momentum - Initial momentum

Momentum is defined as the product of mass and velocity.

Given data:

Mass of the baseball (m) = 0.140 kg

Initial velocity of the baseball ([tex]v_i_n_i_t_i_a_l)[/tex] = 30.0 m/s

Final velocity of the baseball ([tex]v_f_i_n_a_l_[/tex]) = -44.0 m/s (negative sign indicates opposite direction)

To calculate the initial momentum, we multiply the mass by the initial velocity:

Initial momentum = m * [tex]v_i_n_i_t_i_a_l[/tex] = 0.140 kg * 30.0 m/s = 4.2 kg·m/s

To calculate the final momentum, we multiply the mass by the final velocity:

Final momentum = m * [tex]v_f_i_n_a_l_[/tex] = 0.140 kg * (-44.0 m/s) = -6.16 kg·m/s

Now we can find the change in momentum:

Change in momentum = Final momentum - Initial momentum

Change in momentum = (-6.16 kg·m/s) - (4.2 kg·m/s)

Change in momentum = -10.36 kg·m/s

Therefore, the change in momentum for the baseball is -10.36 kg·m/s. The negative sign indicates a change in direction, as the ball is hit back in the opposite direction.

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A simple harmonic oscillator (SHO) has a spring constant of 280 N/m, Total energy of 150 J, and a mass of 4.00 kg. What is its
maximum velocity?
Numerical answer is in units of m/s

Answers

The maximum velocity is 8.66 m/s²

As we know, simple harmonic motion refers to a to-and-fro motion in a periodic manner and spring constant refers to the force required to stretch or compress a spring.

The spring constant for a simple harmonic oscillator is given as 280 N/m, the total energy is 150 J and the mass is 4 kgs. We have to find the maximum velocity of the given simple harmonic motion.

We know that Energy = force x perpendicular distance

In an SHM, energy is in the form of Kinetic Energy. Hence, we use the formula for kinetic energy.

To find the maximum velocity, we will apply the formula for kinetic energy.

Since Kinetic Energy = 1/2 mass x velocity²

Therefore, 150 = 1/2 mass x velocity² ; velocity = 8.66 m/s²

Hence, the maximum velocity for the given system of SHM is 8.66 m/s²

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