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enyata [817]
3 years ago
15

I Need Help With This

Physics
2 answers:
timama [110]3 years ago
5 0

use the equations provided it shouldn't he hard

denpristay [2]3 years ago
4 0

The flux is directly proportional to the area of the loop. You don't need the cos theta part in your formula since we are told the magnetic field is perpendicular to the area of the loop. The area is that of a circle, and it is a function of time:

\Phi_B = BA= B\pi r^2(t)=B\pi(r_0+bt)^2

The last form answers 6a.

The induced electromotive force (EMF) is the negative change of the flux, so you can use a derivative to answer the question 6b:

EMF=-\frac{d\Phi_b}{dt}=\frac{d}{dt}[-(B\pi(r_0+bt)^2]=-2B\pi b(r_0+bt)

(b is treated as a constant, so there is no need to differentiate it). As you can see, the EMF is linear in time, and quadratic in b.

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A is the correct answer

Explanation:

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Consider two point charges located on the x axis: one charge, q1= -11.0 nC, is located at x1= -1.675 m; the second charge, q2= 3
Mekhanik [1.2K]

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Please refer to the figure.

q1 is a negative charge, and q2 and q3 are positive charges. So, the force exerted by q1 on q3 is attractive, and the force exerted by q2 on q3 is repulsive, which means F13 is directed towards left, and F23 is also directed towards left.

F_{13} = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_3}{r_1^2} = \frac{1}{4\pi \epsilon_0}\frac{11\times 10^{-9}\times 47.5\times10^{-9}}{(-1.675 - (-1.18))^2} = 1.92\times 10^{-5}N

F_{23} = \frac{1}{4\pi\epsilon_0}\frac{q_2q_3}{r_2^2} = \frac{1}{4\pi\epsilon_0}\frac{31\times 10^{-9} \times 47.5\times 10^{-9}}{(-1.18 - 0)^2} = 9.5\times 10^{-6}

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3 years ago
A force of 44 N will stretch a rubber band 88 cm ​(0.080.08 ​m). Assuming that​ Hooke's law​ applies, how far will aa 11​-N forc
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Answer:

<em>The rubber band will be stretched 0.02 m.</em>

<em>The work done in stretching is 0.11 J.</em>

Explanation:

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Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

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For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = <em>0.02 m</em>

<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.

potential energy stored = \frac{1}{2}ke^{2}

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