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How could you increase the potential energy of the apple?

VMariaS [17]

By putting an apple up on high ground. (That is 1 example)
This creates a higher gravitational force, and when it falls down it would have a lot of kinetic energy. But, if put on low ground, it wouldn't have enough potential energy to increase the kinetic energy.

7 0

3 years ago

Read 2 more answers

According to newton's second law, if you have an object that is put under acceleration due to a force, how could you reduce the

expeople1 [14]

Answer:

We could reduce the force or increase the mass of the object.

Explanation:

Using the definition of force, according to Newton's second law, (F = ma) the acceleration is directly proportional to the force and inversely proportional to the mass. Therefore, if we want to reduce the acceleration of an object we have two options:

We could reduce the force
We could increase the mass

In both cases a will decrease.

I hope it helps you!

7 0

3 years ago

A hoop and a solid disc are relased from rest

Murljashka [212]

Answer:

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

$K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0$

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

$I_h = m_hR^2$

Let's continue with the energy equations:

$m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}$

Similarly <u>for the solid disk</u> with a moment of inertia of (1/2)mR^2:

$K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}$

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

$v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2$