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The correct answer is (c) C.

EXPLANATION: 

Since the acceleration due to gravity varies inversely to the square of the distance hence the apparent weight (net gravitational force) of the passenger with respect to time will not be a straight line but a curve. In between the earth and the mars, there will be a point where the gravitational force due to both of the bodies will be equal and opposite, hence the apparent weight will zero but never negative. Out of the three curves in the figure, only curve C fulfills this condition.

(b) 0.0027 m/s²   

Explanation: 

When moving in the orbit, the moon balances the force on it by the earth with its uniform circular motion in the orbit resulting in centrifugal force. When stopped for an instant, the force on the moon by the earth does not change, only balancing force is absent. So acceleration at that instant = Force /mass remains the same. Only in the orbit, it is the instantaneous centripetal acceleration. 

(c) ​ 6.4m/s² .

Explanation:

When the moon strikes the earth its center is R/4 above the surface of the earth. So the acceleration of the moon will be calculated at R+R/4 = 5R/4 away from the center of the earth. Since the acceleration is inversely proportional to the square of the distance, the acceleration of the moon at the time of strike  

= g*{R/(5R/4)}² = g*(4/5)² = 16g/25 =160/25 =6.4 m/s².  

An apple is dropped from a tree. 

∴ Initial velocity u = 0 ; Time t = 1.5 s 

a = g = 10 m/s² ; Final velocity v = ? 

v = u + at = 0 + 10 x 1.5 = 0 + 15 = 15 m/s 

Distance covered (s) = ut + 1/2 at² 

= 0 × 1.5 + 1/2 × 10 × 1.5 × 1.5 

= 0 + 5 × 2.25 = 11.25 m

Whenever objects fall towards the earth only under the gravitational force of the earth (with no other forces acting on it) then we say that the objects are in the state of free fall.

Example: An object dropped from a height like a stone on the ground.

When a man moves from earth’s pole to equatorial line, the weight of man will decrease, because on poles the value of acceleration due to gravity is more as compared to the equatorial line.

The mass of a stone is very small, due to which the gravitational force produces a large acceleration in it. Due to large acceleration of stone, we see stone falling towards the earth. The mass of earth is, however, very, very large. Due to the very large mass of the earth, the same gravitational force produces very, very small acceleration in the earth, that it cannot be observed. And hence we do not see the earth rising up towards the stone.

w = mg

(Here m = 10 kg,

g = 10 m/s²)

w = 10 x 10 kg m/s²

w = 100 N

At the top of its path in a projectile, the acceleration is in downward direction.

a = \(\frac{F}{m}\)

This acceleration is the acceleration due to gravity and is directed towards the centre of earth.

Hence, option C is correct.

i) u = 49 m/s, g = 9.8 m/s².

Maximum height = u² /2g = (49 x 49)/(2 x 9.8) = 122.5 m.

ii) Total time to return to the surface of the earth = 2u/g = (2 x 49 )/9.8 = 10 s.

Newton’s law of gravitation states that: Every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

F ∝ \(\frac{1}{R^2}\)

Hence, option D is correct.

Initial velocity, u = 20 m/s 

Final velocity, v = 0 

Acceleration due to gravity, g = -9.8 m/s² 

Height, h = ? 

Using the relation v² = u² + 2gh 

h = 20.4m

(a) In weightlessness state, the major problem faced by astronauts in the spacecraft is eating, drinking water, etc. It occurs because these eatable materials do not stick with plate properly and glass water does not pour.

(b) The space flight for a long time adversely affects the astronauts.

(c) Inside the aircraft, all the objects are in a floating position.

On natural satellite like the moon, the man does not experience weightlessness because the mass of moon (natural satellite) is more and it applies a force of gravity on man. But in artificial satellite, the mass of satellite is less, so it imposes less gravitational force on the object due to which the astronaut experience the weightlessness.