Question

Q1) a) Two objects of masses m1 and m2 , placed at a distance d, exerts a force F on each other.
What happens to the force between them, if:
the masses of both the objects are tripled?
the distance between the objects is doubled?

b) An object is thrown vertically upwards with an initial velocity of 100 ms-1. Find
(i) the time taken to reach maximum height.
(ii) maximum height reached​

Answer 1

Answer

a) 3F

ii)1/2F

b)i) 10 seconds

Explanation: If the mass of the object is tripled means the force will also get 3 times the initial  F as it is directly proportional to the mass of the object.

ii) If the distance is doubled the F will become 1/2 of the initial force.

b) i) Given, u = 100 m/s ,  v = 0 m/s ,  g/a = -10m/s2

(as the object is thrown upward so g value is -ve )

➡Time taken to reach maximum height

t = (v - u)/ a= (0 - 100) / (-10) = 10s

Time taken to reach the maximum height =10 seconds

Hope it helped :)

Answer 2

Answer:

$5 { \times 5y \sqrt{. \sqrt{?} } }^{2}$