Objects that gain elastic energy by being stretched out

Which statements correctly compare the gravitational force with the electrical force? Check all that apply.

skad [1K]

The statements that correctly compare the gravitational force with the electrical force are the following:

-The gravitational force can be attractive.

-The electrical force can be repulsive.

-The electrical force can be attractive.

-Any two objects experience a gravitational force between them.

9 0

3 years ago

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If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration

Nataly_w [17]

Answer:

When we double the angular velocity the maximum acceleration $(a_{max})$ will changes by a factor of 4.

Explanation:

Given the angular frequency $(\omega)$ of the simple harmonic oscillator is doubled.

We need to find the change in the maximum acceleration of the oscillator.

$a_{max}=A\omega^2$

Now, according to the problem, the angular frequency $(\omega)$ got doubled.

Let us plug $\omega=2\times \omega$ . Then the maximum acceleration will be $a_{max'}$

$a_{max}=A\omega^2$

$a_{max'}=A(2\times \omega)^2\\a_{max'}=A\times 4\omega\\a_{max'}=4A\omega$

$a_{max'}=4a_{max}$

We can see, when we double the angular velocity the maximum acceleration will changes by a factor of 4.

6 0

2 years ago

Planet Kling has half the radius and 2 times the mass of the Earth. What is the best estimate for the magnitude of the gravitati

Vlad [161]

Answer:80 m/s^2

Explanation:

6 0

3 years ago

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FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes: