1.find the weight of an object at a height 7400km, above the earth surface . The weight of the object at the surface of the earth is 20N and the radius of the earth is 7400 km.
2.A man's weight when taken at the poles is 700N will his weight remain the same .when measure's at the equator will there be an increase or decrease in his weight.explain in brief.
3.2 objects 'a' and 'b' having masses 200kg and 75kg ,moving with velocity 40km/h and 80km/h
a)which one has the greater inertia?
b)which one has greater momentum?
c)which one will stop first if equal negative acceleration is applied on both?
d)which one will travel greater distance?
e)which one has greater impulse?
{pls i need all the answer as fast as i can and pls forgive and correct if any question pattern is wrong (every step should be explained properly because to complete a exam) }




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Weight = gravitational force between Earth and the object
           = G Me m / d² = G Me m / (Re + h)²

G = gravitational constant
Me = mass of earth
m mass of object
d = distance between center of earth and center of mass of object
Re = radius of earth = 7400 km
h = altitude of object above Earth's surface = 7400 km is same as Re
g = acceleration due to gravity on earth's surface = 9.8 m/sec²

Weight on Earths surface = 20 N = GMe m / Re² 

Weight at 7400km height = Re above Earth's surface = GMe m / (Re + h)²
         =  G Me m / 2² Re² = { G Me m /Re² } / 4
         = 20 N / 4 = 5 N

Radius of Earth at poles is less compared to radius of earth at the equator. Since gravitational attraction force is inversely proportional to Radius², the weight is less at the equator than the weight at poles.


Momentum of a = 200 kg * 40 * 1000m /3600sec = 2222.22 kg m /sec
Momentum of b = 75 kg* 80 * 1000m/3600 sec = 1666.66 kg m /sec

a)  mass tells the inertial. So object a has more inertia.

b) momentum of a = product of mass * velocity  is more than momentum of b

c)  v = u + at    or  t = (v-u)/a  = -u/a        as v =0
   Initial speed of b is more than speed of a.  So time to stop b will be more. a stops earlier.

d) s = (v² - u²)/2a = -u²/2a        , as v = 0
  Initial speed is more for object b. so distance traveled by b will be more.

e) impulse depends on momentum (divided by time to stop).
   If time to stop is same for both objects, then object a has more momentum and
   hence it has more impulse.