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Free_Kalibri [48]
3 years ago
7

A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of

3.30 m/s and rebounds with a speed of 1.60 m/s, determine the following. (a) magnitude of the change in the ball's momentum in kg · m/s (Let up be in the positive direction.)
Physics
1 answer:
pochemuha3 years ago
8 0

Answer:

\Delta p=1.3475\ kg-m/s

Explanation:

The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-

We represent

The ball mass =  m = 275 g = 0.275 kg

Thus it goes to the floor and resurfaces upward.

The ball hits the ground at 3.30 m/s speed that is

u = -3.30 m/s which represents the Negative since the ball hits the ground)

It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)

\Delta p=p_f-p_i

\Delta p=m(v-u)

\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))

\Delta p=1.3475\ kg-m/s

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A force of 44 N will stretch a rubber band 88 cm ​(0.080.08 ​m). Assuming that​ Hooke's law​ applies, how far will aa 11​-N forc
Setler79 [48]

Answer:

<em>The rubber band will be stretched 0.02 m.</em>

<em>The work done in stretching is 0.11 J.</em>

Explanation:

Force 1 = 44 N

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Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = <em>0.02 m</em>

<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.

potential energy stored = \frac{1}{2}ke^{2}

==> \frac{1}{2}* 550* 0.02^{2} = <em>0.11 J</em>

3 0
3 years ago
A 900 kg car speeds up from 30 M/S to 80 m/s how much momentum did the car gain?​
AlexFokin [52]

The change in momentum of the car is 45,000 kg m/s

Explanation:

The change in momentum of an object is given by:

\Delta p = m(v-u)

where

m is the mass of the object

u is its initial velocity

v is its final velocity

For the car in this problem, we have

m = 900 kg

u = 30 m/s

v = 80 m/s

Therefore, the change in momentum is:

\Delta p = (900)(80-30)=45,000 kg m/s

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

#LearnwithBrainly

3 0
3 years ago
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