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AlexFokin [52]
3 years ago
8

A mini rail gun made of a copper rod of mass m and radius r rests on two parallel copper rails that are a distance L apart and o

f length d. A magnetic field of magnitude B is directed perpendicular to the rod and the rails. When current is applied to the parallel copper rails, the rod rolls along the rails without slipping. If the rod starts from rest, what is the speed of the rod as it leaves the rails?
Physics
1 answer:
ira [324]3 years ago
7 0

Answer:

Explanation:

Magnetic Force on the rod  F = Bi L

Work done by this force = F X d

=  Bi Ld

This energy is converted into both rotational and linear kinetic energy

= 1/2 I ω² + 1/2 mv²

= 1/2 x 1/2 m r²ω²+ 1/2 mv²

= 1/4 m v² + 1/2 mv²

= 3/4 mv²

So according to conservation of energy

Bi Ld =  3/4 mv²

v = \sqrt{\frac{4BiLd}{3m} }

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As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change
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Refer to the diagram shown below.

m =  the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A =  the amplitude ( the maximum distance) of the mass from the equilibrium
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The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
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The velocity and acceleration are respectively
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In the equilibrium position,
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At the farthest distance (A) from the equilibrium position,
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In the graphs shown, it is assumed (for illustrative purposes) that
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3 years ago
HOW FAR DOES A UNICYCLE TRAVEL AT A SPEED OF 20 M/S FOR 15 SECONDS?​
astra-53 [7]

Given:-

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Therefore,

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3 years ago
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