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Leona [35]
2 years ago
13

2 kg Use 10 m/s2 for g

Physics
1 answer:
Iteru [2.4K]2 years ago
6 0

Answer:

See below

Explanation:

At point A    the PE = mgh = 2 * 10 * 1 = 20 J

  at point B, all of the PE , 20 J , is converted to Kinetic Energy

KE = 1/2 m v^2

20 = 1/2 (2)(v^2 )

20 = v^2     v = sqrt 20 = 4.47 m/s

for the friction part

 vf = vo t  + 1/2 a t^2      vf = final velocity = 0 (stopped)

                                       vo = original velocity = 4.47 m/s  

                                         a = -1 m/s^2

  0 = 4.47 t + 1/2 (-1) t^2

            - .5t^2  + 4.47 t = 0

                 t ( -.5t+ 4.47) = 0    shows t =  4.47/.5 = 8.9 seconds

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Answer:

"A pendulum swinging back and forth" is an example of harmonic motion

X = Xo cos ω t

Explains the back and forth motion of the pendulum

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2 years ago
The circumference of an orbit for a toy on a string is 18 m and the centripetal force is 12 N. Does the centripetal force do any
Snezhnost [94]

Answer:

Work done is 0.

Explanation:

Given that,

The circumference of an orbit for a toy on a string is 18 m, r = 18 m

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5 0
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6 0
3 years ago
The phases of the moon depend on how much of the lighted side of the moon can be seen from earth.is this true or false
Sindrei [870]
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6 0
3 years ago
A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec
Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

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So angular frequency \omega =2\pi f=2\times 3.14\times 8=50.24rad/sec

Velocity of the string wave is equal to v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec

Power of wave propagation is equal to P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt

So power of the wave will be equal to 5.464 watt

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3 years ago
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