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Leto [7]
3 years ago
6

A bowling ball of mass 5 kg hits a wall going 7 m/s and rebounds at a speed of 2 m/s. What was the impulse applied to the bowlin

g ball
Physics
2 answers:
LekaFEV [45]3 years ago
6 0
The impulse applied to the bowling ball is 45 kg m/s
gizmo_the_mogwai [7]3 years ago
3 0

Explanation :

Given that,

Mass of the ball, m = 5 kg

Initial velocity of the ball, u = 7 m/s

Final speed of the ball, v = 2 m/s

We know that the impulse applied to an object is equal to the change in momentum.

J=\Delta p

J=p_f-p_i

J=m(v-u)

J=5\ kg(2\ m/s-7\ m/s)

J = -25 kgm/s

Hence, this is the required solution.

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no its not like the undertow in the ocean

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Which item could you use in place of an ammeter to demonstrate that a
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Answer:

the answer is D

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2 years ago
A 5c charge experiences a force of 40n when put at a certain location in space. What is the electric field at that location?
stepladder [879]
The 'strength' of the electric field is the force on 1C of charge at that point.

At this 'certain location', the field is 40/5 = 8 newtons per coulomb = <u>8 volts</u>
3 0
3 years ago
A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di
Lyrx [107]

Answer:

D

Explanation:

From the information given:

The angular speed for the block \omega = 50 \ rad/s

Disk radius (r) = 0.2 m

The block Initial velocity is:

v = r \omega \\ \\  v = (0.2  \times 50) \\ \\  v= 10 \ m/s

Change in the block's angular speed is:

\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s

However, on the disk, moment of inertIa is:

I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2

The time t = 10s

∴

Frictional torques by the wall on the disk is:

T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10})  \\ \\ =0.6 \ N.m

Finally, the frictional force is calculated as:

F = \dfrac{T}r{}

F= \dfrac{0.6}{0.2} \\ \\ F = 3N

8 0
3 years ago
You stand on a frictionless platform that is rotating with an angular speed of 5.1 rev/s. Your arms are outstretched, and you ho
timurjin [86]

Answer:

\omega'=19.419\ rev.s^{-1}

Explanation:

Given:

angular speed of rotation of friction-less platform, \omega=5.1\ rev.s^{-1}

moment of inertia with extended weight, I=9.9\ kg.m^2

moment of inertia with contracted weight, I'=2.6\ kg.m^2

<u>Now we use the law of conservation of angular momentum:</u>

I.\omega=I'.\omega'

9.9\times 5.1=2.6\times \omega'

\omega'=19.419\ rev.s^{-1}

The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.

5 0
3 years ago
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