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Leviafan [203]
3 years ago
10

29. A chain is wrapped around a pulley and pulled

Physics
1 answer:
kati45 [8]3 years ago
6 0

Explanation:

use torque =I\alpha

and obtain alpha by \alpha =(17×2×3.14)/60×5

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What formula gives the strength of an electric field, E, at a distance from a known source charge?
Deffense [45]
<h2>Hello!</h2>

The answer is: Coulomb's law equation.

<h2>Why?</h2>

The Coulomb's law states that the strength of an electric field (between two charges) can be calculated by multiplying their charges and dividing it into the square of the distance between their centers.

E=\frac{k*q*Q}{d^{2} }

Where:

E = Electric Field Strenght

k=9.0*10^{9}  \frac{N.m^{2} }{C^{2} }

q=TestCharge\\Q=SourceCharge\\

d = separation between charges (m)

Have a nice day!

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4 years ago
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Answer:

B. v = λ X f

<em><u>Hope this helps! </u></em>

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3 years ago
Which of the following is a true statement? A. Electromagnetic waves consist of only changing electric fields. B. Electromagneti
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C. Electromagnetic waves don't always need a medium to travel. Note that they do vary in wavelength and frequency however their speed is fixed. Also, EM waves are always transverse and they consist of vibrating electric and magnetic fields.
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3 0
3 years ago
Juan is on a morning jog. His speed is represented in the graph. At what rate of speed is Juan running between 4min and 6min? Ac
Nesterboy [21]

Answer:

<em>a) The speed of Juan is 250 m/min </em>

<em>b) In that interval, he's at rest </em>

<em>c) Juan would have needed 12 minutes to run 3000 m</em>

Explanation:

<u>Distance vs Time Graph Analysis </u>

If time is on the horizontal axis, and the distance is in the vertical axis, then the slope of the graph represents the instantaneous speed of the moving object. The graph shows three different zones which reflect Juan's jogging at that morning.

a) The period between 4 and 6 minutes clearly shows Juan is moving in such a way the distance increases when time passes. The speed of the motion can be obtained by computing the slope of the line. We can locate the points (4,1000) and (6,1500). We compute the speed  

\displaystyle v=\frac{1500-1000}{6-4}=\frac{500}{2}=250 m/min

The speed of Juan is 250 m/min

b) We can see between 7 and 11 minutes, Juan's distance is not changing because he stopped running. In that interval, he's at rest

c) We have already determined the speed on the first 7 minutes (the same as between 4 and 6 minutes). We know that

\displaystyle v=\frac{x}{t}

Where v is the speed, x is the distance, and t is the time

We need to know the time needed to travel x=3000 m at the initial speed v=250 m/min, so we solve the equation for t

\displaystyle t=\frac{x}{v}

\displaystyle t=\frac{3000}{250}=12\ minutes

Juan would have needed 12 minutes to run 3000 m

3 0
4 years ago
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