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Sav [38]
3 years ago
12

Please view image attached!

Physics
1 answer:
Marat540 [252]3 years ago
8 0

Answer:

you would expect a change in the graph

Explanation:

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What are some ways that electric fields are similar to magnetic fields?
sattari [20]

Answer:

The correct answers are the proportionality of the fields concerning distance, vector fields, and forces at a distance.

Explanation:

The similarities between magnetic fields and electric fields are that electric fields are produced by two charges that can be positive and negative. Magnetic fields are associated with two magnetic poles, although they are also produced by moving charges. Both fields are inversely proportional to the square of the distance between the sources, both fields are vectorial and both act by distant forces.

Have a nice day!

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3 years ago
I want ti know how to study​
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Answer:

Make sure everything is organized have a planner it can help

Get rid of all distractions

Listen to music if it helps you concentrate

Have your notes

Being willing to stay focus on what you are doing

Understand what you are doing

And most off all Be Happy and Remain Calm : )

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4 years ago
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The three types of seismic waves produced by an earthquake are primary, secondary, and
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<span>The three types of seismic waves produced by an earthquake are primary, secondary, and (D.) surface.</span>
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A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho
Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg)  = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

         = (500)(0.5)(0.1 × 10⁻⁶)

         = 2.5 × 10⁻⁵A

C)  If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}

= 1250W

d) Final peak=

P= Ik/e

= = P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W

P = 2.5 × 10⁷W

5 0
3 years ago
Read 2 more answers
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