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Veseljchak [2.6K]
3 years ago
11

The wheel of a car has a radius of 0.413 m. If the wheel rotates at 456 revolutions per minute, what is the speed (in m/s) at wh

ich the outer edge of the wheel is moving
Physics
2 answers:
suter [353]3 years ago
7 0

Answer: The speed of the edge of the wheel = 19.72m/s

Explanation:

The liner speed of the edge of the wheel is given as;

V=wr

V= linear speed =?

w= angular velocity =456rpm

Converting rpm to rad/s

1rpm = π/30 rad/s

456rpm = 47.75 rad/s

r=0.413m

V=47.75 × 0.413

V= 19.72m/s

Therefore speed of the edge of the wheel = 19.72m/s

irakobra [83]3 years ago
6 0

Answer:

The speed (in m/s) at which the outer edge of the wheel is moving is 19.722 m/s

Explanation:

the solution is in the attached Word file

Download docx
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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.

8 0
3 years ago
Which of these best describes heat transfer through conduction? A) Heat transfer as electrons are excited in atoms. B) Heat tran
Setler79 [48]
I'd go for D here. It also fits in with the idea of thermal expansion - as something is heated up, molecules vibrate and maybe collide. they vibrate with bigger amplitudes, so taking up more space, so expanding. maybe
7 0
3 years ago
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Which statement describes a good physical property of copper
frez [133]
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4 0
3 years ago
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If 270 watts of power is used in 42 seconds, how much work was done<br>​
navik [9.2K]

Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

<u>Step 1:</u><em><u> Find the Formula</u></em>

P = W / T

<em><u> </u></em>

<u>Step 2: </u><u><em>Make W the subject of the equation.</em></u>

W = PT

<u>Step 3:</u><u> </u><u><em>Given.</em></u>

P = 270 Watts, T = 42 seconds

<u>Step 4:</u><u><em> Substitute these values into equation 2 .</em></u>

W = 270(42)

<u>Step 5:</u><u> </u><u><em>Simplify.</em></u>

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

4 0
4 years ago
You move a 2.5 kg book from a shelf that is 1.2 m above the ground to a shelf that is 2.6 m above the ground. What is the change
Sophie [7]
The change in potential energy of an object is given by
U=mg \Delta h
where
m is the mass of the object
g is the gravitational acceleration
\delta h is the increase in altitude of the object

In our problem, m=2.5 kg is the mass of the book, g=9.81 m/s^2 and 
\Delta h=2.6 m -1.2 m=1.4 m is the increase in altitude of the book, so its variation of potential energy is
U=mg\Delta h=(2.5 kg)(9.81 m/s^2)(1.4 m)=34.3 J
8 0
4 years ago
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