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mafiozo [28]
3 years ago
6

In a lab experiment a light flexible string is wrapped around a solid cylinder with mass M = 9.50 kg and radius R. The cylinder

rotates with negligible friction about a stationary horizontal axis. A mass m = 3.00kg is tied at the free end of the string and release and the mass moves downward. Find the magnitude of the acceleration of the object of mass m.

Physics
1 answer:
Lunna [17]3 years ago
5 0

Answer:3.79 m/s^2

Explanation:

Given

Mass M=9.5 kg

m=3 kg

Net Force is equivalent to \sum F=ma

with tension T in the string    

For mass m

mg-T=ma

T=mg-ma--------1

For cylinder

T\cdot R=I\times \alpha

I for solid cylinder is \frac{2}{5}MR^2 , and \alpha =\frac{a}{R}

thus T=\frac{Ma}{2}----2

Substitute the value of T we get

\frac{Ma}{2}=mg-ma

a(\frac{M}{2}+m)=mg

a=\frac{mg}{\frac{M}{2}+m}

a=3.79 m/s^2

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Answer:

c. 307 nm

Explanation:

angular position of first dark fringe =  λ / d ,   λ is wavelength and  d is width of slit .

(40 x π ) / 180 = 410 / d

angular position of second  dark fringe =  2 x λ / d ,   λ is wavelength and  d is width of slit .

(60 x π ) / 180 = 2 x λ / d

Dividing these equations

60 / 40 =  2 x λ / 410

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A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the
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Answer:

c. V = 2 m/s

Explanation:

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Also we know that:

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Also, the moment of inertia of the disk is equal to:

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Answer:

The statement is not correct.

Explanation:

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