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2 years ago

Which of the following objects is in static equilibrium?

Physics

1 answer:

Likurg_2 [28]2 years ago

4 0

<h3><u>Answer:- </u></h3>

$\longrightarrow \textsf{D. A bicycle sitting on the ground}$

<h3><u>Solution</u>:-</h3>

<u>Let us check all options -</u>

A) An apple falling from a tree is under the influence of gravity. Thus some force is acting on the apple and it can not said to be in state of equilibrium.

B) A car moving at a constant velocity may said to be in state of equilibrium but the since the car is in motion ,equilibrium is dynamic in nature .Thus , it is not in static equilibrium.

C) A motorcycle speeding up has some acceleration due to some forces. Thus it can also not be said in state of static equilibrium.

D) Since the force acting on bicycle sitting down in zero as well as it is in state of rest , thus bicycle sitting on the ground may said to be in static equilibrium.

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Answer:

c. $4.582\times10^{21} kg$

Explanation:

$r_{i}$ = Initial distance between asteroid and rock = 7514 km = 7514000 m

$r_{f}$ = Final distance between asteroid and rock = 2823 km = 2823000 m

$v_{i}$ = Initial speed of rock = 136 ms⁻¹

$v_{f}$ = Final speed of rock = 392 ms⁻¹

$m$ = mass of the rock

$M$ = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

$(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg$

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