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NARA [144]
3 years ago
7

The fact that the magnetic field generates a force perpendicular to the instantaneous velocity of the particle has implications

for the work that the field does on the particle. As a consequence, if only the magnetic field acts on the particle, its kinetic energy will ____________.
__ increase over time
__ decrease over time
__ remain constant
__ oscillate
Physics
1 answer:
Amanda [17]3 years ago
4 0

Answer:

The kinetic energy will remain constant

Explanation:

Charged particles in a magnetic field feel a force perpendicular to their velocity,  since the force is F = qvB, and a charged particle feels a force of constant magnitude always directed perpendicular to its motion which causes circular motion in the magnetic field but the speed and kinetic energy of the particle will remain constant.

Therefore, if only the magnetic field acts on the particle, its kinetic energy will remain constant.

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In a real system of levers, wheel or pulleys, the AMA (actual mechanical advantage) is less than the IMA (ideal mechanical advantage) because of the presence of friction.

In fact, the IMA and the AMA of a machine are defined as the ratio between the output force (the load) and the input force (the effort):
IMA= \frac{F_{out}}{F_{in}}
however, the difference is that the IMA does not take into account the presence of frictions, while the AMA does. As a result, the output force in the AMA is less than the output force in the IMA (because some energy is dissipated due to friction), and the AMA is less than the IMA.
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3 years ago
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A 62.0 kg sprinter starts a race with an acceleration of 1.44 m/s2. If the sprinter accelerates at that rate for 30 m, and then
Dominik [7]

Answer:

The time taken for the race is 17.20 s.

Explanation:

It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

v^{2}-u^{2}= 2as

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

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v^{2}-(0)^{2}= 2(1.44)(30)

v= 9.3 m^{-1}

Calculate time taken to cover 30 m distance.

The expression for the equation of motion is as follows;

s=ut+\frac{(1)}{2}at^2

Put u= 0, s=30 m and a= 1.44 ms^{-2}.

30=(0)t+\frac{(1)}{2}(1.44)t^2

t=6.45 s

Calculate the time taken to complete his race.

T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

T=6.45+\frac{s'}{v}

Put s= 30 m, v= 9.3 m^{-1} and s'= 100 m.

T=6.45+\frac{(100)}{9.3}

T= 17.20 s

Therefore, the time taken for the race is 17.20 s.

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