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Trava [24]
3 years ago
15

A machine part has the shape of a solid uniform sphere of mass 205 g and diameter 4.10 cm. It is spinning about a frictionless a

xle through its center, but at one point on its equator, it is scraping against metal, resulting in a friction force of 0.0200 N at that point.
(a) Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation.
(b) How long will it take to decrease its rotational speed by 22.0 rad/s?

Physics
1 answer:
kondor19780726 [428]3 years ago
3 0

Answer:

Answer is in the following attachment.

                                                                                                                                                                                                 

Explanation:

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HELP PLEASE<br> 60 POINTS <br> HAVE A GREAT REST OF YOUR DAY PEOPLE :&gt;
igor_vitrenko [27]

Answer:

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Explanation:

I hope this helps!

4 0
3 years ago
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A sprinter starts from rest and reaches a speed of 15 m/s in 4.25 s. Find his acceleration
kakasveta [241]

Answer:

a=3.53 m/s^2

Explanation:

Vo=0 m/s (because he is not moving at the start)

V1=15 m/s

t= 4.25 s

a = (V1-Vo) / t = 15/4.25 = 3.53 m/s^2

5 0
3 years ago
A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to
jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire  =3.7\ MA

b) Current at the center of the circular coil =2.48\times 10^{5}\ A

c) Current near the center of a solenoid 236.8\ A

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}

\mu=4\pi \times 10^{-7}\ \frac{henry}{m}

Plugging the values,

Conversion 1\times 10^6 A = 1\ MA,and 2cm=\frac{2}{100}=0.02\ m

I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA

⇒Magnetic Field at the center due to circular coil (at center) is given by,B=\frac{\mu\times I (N)}{2(a)}

So I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A

⇒Magnetic field due to the long solenoid,B=\mu\ nI=\mu (\frac{N}{l})I

Then I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A  

So the value of current are  3.7 MA,2.48\times 10^{5} A and 236.8\ A respectively.

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Naya [18.7K]

guaranteeing freedom of speech, press, assembly, and exercise of religion

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Which of the following foods contains large amounts of protein? chicken, beans, fish, dairy, berries
d1i1m1o1n [39]

Answer: chicken hope this helps!

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