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ella [17]
3 years ago
7

A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnit

ude of the impulse exerted on the ball by the floor
Physics
1 answer:
gayaneshka [121]3 years ago
4 0

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns

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Part A.)Six boxes held at rest against identical walls.
krok68 [10]

Answer:

Explanation:

When a body is held against a vertical wall , to keep them in balanced position , normal force is applied on their surface . this force creates normal reaction which acts against the normal force and it is equal to the normal force as per newton's third law . Ultimately friction force is created which is proportional to normal force and it acts in vertically upward direction . It prevents the body from falling down .

Hence normal force = reaction force .

From second law also net force is zero , so if  normal force is N and reaction force is R

R - N = mass x acceleration = mass x 0 = 0

R = N .

Ranking normal force from  highest  to smallest

150 N , 130 N , 120 N

B )

Frictional force is equal to the weight of the body because the body is held at rest .

Ranking of frictional force form largest to smallest

7 kg , 5 kg , 3 kg , 1 kg .

Here frictional force is irrespective of the normal force acting on the body  because frictional force adjusts itself so that it becomes equal to weight in all cases here because it always balances the weight of the body .

6 0
3 years ago
A ball is thrown up at a speed of 20 m/s.
ioda

Given:

(Initial velocity)u=20 m/s

At the maximum height the final velocity of the ball is 0.

Also since it is a free falling object the acceleration acting on the ball is due to gravity g.

Thus a=- 9.8 m/s^2

Now consider the equation

v^2-u^2= 2as

Where v is the final velocity which is measured in m/s

Where u is the initial velocity which is measured in m/s

a is the acceleration due to gravity measured in m/s^2

s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.

Substituting the given values in the above formula we get

0-(20x20)= 2 x- 9.8 x s

s= 400/19.6= 20.41m

Thus the maximum height attained is 20.41 m by the ball

6 0
3 years ago
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