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slava [35]
3 years ago
14

a 4.0 kilogram ball moving at 8.0 m/s to the right collides with a 1.0 kilogram ball at rest. after the collision the 4.0 kilogr

am ball moves at 4.8 m/s to the right. what is the velocity of the 1 kilogram ball?
Physics
1 answer:
Inga [223]3 years ago
7 0
Step by step equation
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oksian1 [2.3K]

A. because it's been repeated multiple times

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ANSWER ASAP!!!If a wave machine produces 10 waves in 2 seconds, what is the frequency of the machine in Hertz?
Rufina [12.5K]

Answer:

5Hz

Explanation:

Frequency = No. of waves per second

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(ie) 5 waves are produced in 1 second

Therefore the frequency of wave is 5 Hz

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3 years ago
A ball is launched from ground level at 20 m/s at an angle of 40° above the
DedPeter [7]

(a) The ball's height <em>y</em> at time <em>t</em> is given by

<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :

0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )

<em>t</em> = 0   or   (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 <em>g t</em>

<em>t</em> = (40 m/s) sin(40º) / <em>g</em>

<em>t</em> ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So

0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>

where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :

<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)

<em>y</em> ≈ 8.4 m

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Answer:

hope fully it help s

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The Red Sea is widening at a rate of 1.25 centimeters per year. How many years will it take to widen another 812.5 centimeters?
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