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morpeh [17]
3 years ago
6

A cyclotron operates with a given magnetic field and at a given frequency. If R denotes the radius of the final orbit, the final

particle energy is proportional to:
Physics
1 answer:
Archy [21]3 years ago
5 0

Answer:

the final particle energy is proportional to

Emax \infty\ R^2

Explanation:

The computation of the final particle energy is as follows;

As we know that

Energy is respect with the radious as

Emax = \frac{q^2R^2B^2}{2m}

In the case when other quantities would remains the same or constant

So, the final particle energy is proportional to

Emax \infty\ R^2

The same is to be considered and relevant too

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When an object is falling and reaches a constant velocity, the net force on the object is ____ and the weight of the object is e
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When an object is falling and reaches a constant velocity, the net force on the object is <em>zero</em> (it's not accelerating), and the weight of the object is equal to <em>the force of air resistance against the object</em>.  (choice-D)

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A jet, sitting on the runway, takes off and accelerates at 8.0 m/s for 16s. How far did the jet travel down the runway?
nydimaria [60]

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2.4 m/s". 1

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8 0
3 years ago
Frank’s automobile engine runs at 100∘C. One day, when the outside temperature is 15∘C, he turns off the ignition and notes that
lapo4ka [179]

Answer:

the engine cool to 40^{o}C at 14.07 minutes

Explanation:

Given information

T(5) = 70^{o}C

T_{0} = 100^{o}C

C = 15^{o}C

Newton's law of cooling :

T(t) = C + (T_{0} - C) e^{-kt}

where

T(t) = temperature at any given time

C = surrounding temperature

T_{0} = initial temperature of heated object

k = cooling constant

to find the the time when the engine will be cooled down to 40^{o}C, we first need to find the cooling constant, k

when t = 5, T(5) = 70^{o}C

so,

T(t) = C + (T_{0} - C) e^{-kt}

T(5) = 15 + (100 - 15) e^{-5k}

70 = 15 + (85) e^{-5k}

e^{-5k} = (70 - 15) / 85

-5k = ln (55/85)

k = - ln (55/85) / 5

k = 0.087

thus, we have the eqaution

T(t) = 15 + (85) e^{-0.087t}

now we can determine the time when T(t) = 40^{o}C

40 = 15 + (85) e^{-0.087t}

e^{-0.087t} = (40-15)/85

-0.0087t = ln (25/85)

t = - ln (25/85)/0.087

t = 14.07 minutes

8 0
3 years ago
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