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GarryVolchara [31]
2 years ago
10

An electron moving north encounters a uniform magnetic field if the magnetic field points east what is the direction of the magn

etic force on the electron
Physics
1 answer:
Alex787 [66]2 years ago
4 0

Answer:

force is leaving the sheet of paper

Explanation:

The magnetic force is given by the expression

       F = q v x B

where the boldface indicates vectors, this expression can be separated in its module

       F = q v B sin θ

dodne θ at the angle between velocity (v) and the magnetic field (B)

the direction can be found by the right-hand rule, where for a positive charge the thumb points in the direction of speed, the other fingers extended in the direction of the magnetic field and the palm in the direction of force, if the charge is negative the direction of the force contrary to the direction of the palm.

Then

the thumb is in the North direction

fingers extended in the east direction

The palm points entering the sheet, but since the charge is negative the force is leaving the sheet of paper

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Which of the following statements best describes the characteristic of the restoring force in the spring-mass system described i
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Answer : The restoring force is directly proportional to the displacement of the block.

Explanation :

Restoring force is defined as the force that is exerted by the spring due to its mass.

Mathematically, the restoring force can be written as :

F\propto-x

F = - k x

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x is the displacement caused due to the mass.

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So, it is clear that the restoring force is directly proportional to the displacement of the block.

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5 0
2 years ago
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You will have to use this formula:
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Final Velocity (V) = 4m/s
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Then:

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Ps: It's value is negative because the she was in retrograde motion.

Answer: Her acceleration is -2 m/s^2.
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2 years ago
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A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
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We use the kinematic equations,

v=u+at                                          (A)

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

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

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2 years ago
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~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey
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