3 years ago

What is the net magnetic flux through any closed surface?

Physics

1 answer:

Paul [167]3 years ago

3 0

Answer:

The net magnetic flux through any closed surface must always be zero.

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wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the

lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

$E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)$

The electric field due to charge QB at P is EB leftwards

$E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)$

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

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3 years ago

An airplane starts from A and flies to B at a constant speed. After reaching B it returns to A at the same speed. There was no w

Dafna1 [17]

Answer:

When there is wind it takes longer

Explanation:

With no wind, the round trip time is

$t_1=\frac{d}{v}+\frac{d}{v}=\frac{2d}{v}$

When we have a constant wind speed w

$t_2=\frac{d}{v-w} +\frac{d}{v+w} =\frac{2vd}{v^{2} -w^{2}}$

comparing the reciprocal times;

$\frac{1}{t_2}=\frac{v^{2}-w^{2} }{2vd}=\frac{v}{2d}-\frac{w^{2}}{2vd} \leq \frac{v}{2d}=\frac{1}{t_1}$

This means that t1 is smaller than t2, ergo, it takes longer with wind

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