Сс Explain the effect of Earth's gravity on objects on the surface of Earth.

Physics

1 answer:

dem82 [27]3 years ago

6 0

Answer:

The effect of gravity on object makes objects stable and not floating in the air . It also makes any object thrown up to return back

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The Atwood’s machine shown consists of two blocks of mass m1 and m2 that are connected by a light string that passes over a pull

Talja [164]

(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

<h3 /><h3>The given parameters:</h3>

Mass of block 1 = m1
Mass of block 2, = m2
Height of block 1 above the ground, = h1
Height of block 2 above the ground = h2

The total initial mechanical energy of the two block system is calculated as follows;

$m_1gh_1 + \frac{1}{2} m_1v_1_i^2 = m_2gh_2 + \frac{1}{2} m_2v_2_i^2\\\\m_1gh_1 + 0 = m_2gh_2 + 0\\\\m_1gh_1 = m_2gh_2\\\\m_1gh_1 - m_2gh_2 = 0$

When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;

$m_1g(h_1 + h_2) + K.E_1 = \frac{1}{2}m_2v_{max}^2 + P.E_2\\\\m_1g(h_1 + h_2 ) -PE_2 = \frac{1}{2}m_2v_{max}^2 - K.E_1\\\\m_1g(h_1 + h_2 ) - 0= \frac{1}{2}m_2v_{max}^2 - 0\\\\m_1g(h_1 + h_2 ) = \frac{1}{2}m_2v_{max}^2\\\\W = \frac{1}{2}m_2v_{max}^2$

Thus, we can conclude the following before the block m2 reaches the ground;

For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

5 0

3 years ago

What is the source of energy

liberstina [14]

Fossil Fuels is the largest source for energy

Hope this is what you were asking :)

8 0

3 years ago

A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally

dalvyx [7]

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

$F_{C} = F_{G}$