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nikdorinn [45]
3 years ago
15

There is an analogy between rotational and linear physical quantities. what rotational quantities are analogous to distance and

velocity?
Physics
2 answers:
Scrat [10]3 years ago
8 0

Answer:

Rotational units can be said to be analogous to linear physical units, for example, linear distance is analogous to angular displacement, since angular displacement is defined as the distance a body travels in a rotary motion.

On the other hand, linear velocity is analogous to angular velocity, since said angular velocity is the angular distance that a body covers per unit of time.

Explanation:

SVEN [57.7K]3 years ago
5 0

ANSWER:

Angular displacement is analogous to distance.

Angular velocity is analogous to velocity.

EXPLANATION:

FOR  ANGULAR DISPLACEMENT AND DISTANCE

Angular displacement = theta

theta = s/r

s = linear dispalcement/distance

r= radius.

so ,

linear displacement/distance  s = theta * r.

FOR  ANGULAR VELOCITY AND VELOCITY

Angular velocity = w

w = v/r

v = velocity

r = radius

so,

velocity v = w * r.

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Two teams are playing tug of war. Team A pulls to the right with a force of 450 N. Team B pulls to the left with a force of 415
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35 N to the right.

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3 years ago
A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig
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Answer:

The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

Explanation:

From the question we are told that

The distance between earth and Retah is  d = 20 \ light \ hours =  20 * 3600 *  c =  72000c \ m

Here c is the peed of light with value c =  3.0*10^8 m/s

The time taken to reach Retah from earth is  t =  25 \ hours  =  25 * 3600 =90000 \ sec

The velocity of the spacecraft is mathematically evaluated  as

     v_s =  \frac{d }{t}

substituting values

   v_s =  \frac{72000 * 3.0*10^{8} }{90000}

    v_s =  2.40*10^{8} \ m/s

The time elapsed in the spacecraft’s frame is mathematically evaluated as

      T  =  t *  \sqrt{ 1 -  \frac{v^2}{c^2} }

substituting value

       T  =  90000 *  \sqrt{ 1 -  \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }

        T = 54000 \ s

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So  The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

       

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