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alex41 [277]
3 years ago
9

Consider the example of a positive charge q moving in the negative z-direction with speed v with the local magnetic field of mag

nitude B in the z-direction.
Find F, the magnitude of the magnetic force acting on the particle. Express your answer in terms of v, q, B, and other quantities given in the problem statement.
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
4 0

Answer:

<em>F</em> = <em>0</em>

Explanation:

This problem can be solved by using the equation for the force produced by a magnetic field over a charge

<em>F</em> = q<u><em>v</em></u> X <em>B</em>

That is, the cross product between velocity and the magnetic field.

In this case, v is in the negative z-direction <em>v</em> = -v k^,

and the magnetic field is in the positive z-direction <em>B</em> = z k^

Hence, the cross product is

F = q<em>v </em>X <em>B</em> = 0i^ - 0j^ + 0k^ = <em>0</em>

In this case the net force is zero due to both velocity and magnetic field vectors are in the z-direction

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Lina20 [59]
Work is force*displacement if the force and displacement is parallel. 


a. You can average the force over the distance so W = Fave*d 

<span>b The force part of that multiplication is zero. </span>

<span>c. You can form the average force for the interval from 2 to 3 and find the work for that section and then consider the interval from 3 to 4, find the work and add the 2 work results.


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3 years ago
Calculate the potential energy stored in an object of mass 50 kg at a height of 20 m from the ground.
bearhunter [10]

Answer:

potential energy=mgh

=50×10×20

=10000 J

8 0
3 years ago
Read 2 more answers
if the current through a resistor is increased by a factor of 4, how does this affect the power dissipated?
lbvjy [14]

The new current will be 4I. The power dissipated by the resistor will increase by a factor of 16.

<h3>What is a resistor?</h3>
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4 0
1 year ago
Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (
pentagon [3]

Answer:

q=2.997\times 10^{-4}C

Sign-Negative

Explanation:

We are given that

Electric field =E=100NC^{-1} (Radially downward)

Acceleration=0.19 ms^{-2}(Upward)

Mass of charge=3 g=3\times 10^{-3}kg

1kg=1000g

We have to find the magnitude and sign of  charge would have to be placed on a penny .

By newton's second law

\sum F_y=ma

\sum F_y=qE-mg

Substitute the values then we get

qE-mg=ma

Substitute the values then we get

q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)

100q-29.4\times 10^{-3}=0.57\times 10^{-3}

100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}

q=\frac{29.97\times 10^{-3}}{100}

q=2.997\times 10^{-4}C

Sign of charge =Negative

Because electric force acting  in opposite direction of electric field therefore,charge on penny will be negative.

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3 years ago
2. Use physics terms to explain the benefits of crumple zones in modern cars.
Elden [556K]

Answer:

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Explanation:

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