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frozen [14]
3 years ago
14

slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of

25 m/s and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?
Physics
1 answer:
Ivan3 years ago
6 0

Answer: 800N

Explanation:

Given :

Mass of ball =0.8kg

Contact time = 0.05 sec

Final velocity = initial velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is can be calculated using the relation;

Force(F) = mass(m) * average acceleration(a)

a= (initial velocity(u) + final velocity(v))/t

m = 0.8kg

u = v = 25m/s

t = contact time of the ball = 0.05s

Therefore,

a = (25 + 25) ÷ 0.05 = 1000m/s^2

Therefore,

Magnitude of average force (F)

F=ma

m = mass of ball = 0.8

a = 1000m/s^2

F = 0.8 * 1000

F = 800N

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Explanation:

Let's consider orbital mechanics. To get an object in orbit, we need it to fall to earth parallel to the earth's surface. To understand it easily imagine a projectile thrown horizontally further and further away, at one point, the projectile hits the cannon from behind. Considering there is no wind resistance, that would be a projecile in orbit.

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A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
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Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

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Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

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