Physics, asked by nidhibharuka, 1 month ago

1. A 500 g ball is moving with a speed of 50 m/s. It is brought to rest in 0.05 s. Calculate the change in momentum of the ball and the average force applied.
2. A body is thrown vertically upwards from the top of a tower of height 54.2m with an initial velocity of 21.6 m/s. Taking g as 10 m/s2 , calculate :-(i)the height to which it will rise before returning to the ground, (ii) the velocity with which it will strike the ground and (iii) the total time of the journey.
3. A square plate of side 2x 103 cm is placed horizontally 100 cm below the surface of water. Calculate the total thrust on the plate. Given:- Atmospheric pressure= 1.013x105 Pa, density of water= 1000 kg m-3, g= 9.8 m s-2.
4. The areas of pistons in a hydraulic machine are 5cm2 and 10 cm2 respectively. What force on the smaller piston will support a load of 150 N on the larger piston?
5. At a given place, a mercury barometer records a pressure of 0.70 m of Hg. What would be the height of water column if mercury in barometer is replaced by water? Given density of mercury is 13600 kg m-3.

Answers

Answered by prithigathiyagarajan
0

nono no no no nononononono ononono

Answered by RitaNarine
0

1. Given,

Mass of the ball, m = 500 g = 0.5 kg

Initial velocity, u = 50 m/s

Final velocity, v = 0 m/s

Time to bring the ball to rest, t = 0.05 s

To find,

The change in momentum of the ball, \DeltaP = ?

the average force applied, F= ?

Solution,

The change in momentum,

\DeltaP = m(v-u)

     = 0.5(0-50)

     = -25

[The negative sign indicates that the change in momentum is in the opposite direction of the motion]

Hence, the change of momentum of the ball is -25 kg m/s.

Now, the average force applied,

F = m*(\frac{v-u}{t} )\\F = 0.5*(\frac{0-50}{0.05} )\\F = - 500 N

[The negative sign indicates that the force acts in the opposite direction of the motion]

Hence, the average force applied is -500 N.

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