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Snowcat [4.5K]
3 years ago
8

The 75.0 kg hero of a movie is pulled upward with a constant acceleration of 2.00 m/s2 by a rope. What is the tension on the rop

e?
585N


75.0N


885N


11.8N
Physics
1 answer:
lutik1710 [3]3 years ago
7 0

Use Newton's second law. The net force on the hero is

∑ <em>F</em> = <em>T</em> - <em>m g</em> = <em>m</em> (2.00 m/s²)

where

• <em>T</em> = tension in the rope

• <em>m</em> = 75.0 kg = mass of the hero

• <em>g</em> = 9.80 m/s² = acceleration due to gravity

Solve for <em>T</em> :

<em>T</em> = (75.0 kg)<em> </em>(<em>g</em> + 2.00 m/s²) = 885 N

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qV=\frac{1}{2}mv^2

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v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

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F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

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E)

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