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butalik [34]
3 years ago
6

The earth's radius is 6.37×10^6m; it rotates once every 24 hours. with the angular speed of 7.3 x 10^-5 what is the speed of a p

oint on the earth's surface located at 2/3 of the length of the arc between the equator and the pole, measured from equator? (Hint: what is the radius of the circle in which the point moves?)
Physics
1 answer:
alexira [117]3 years ago
6 0
The speed of the earth's surface located at 2/3 of the length of the arc between the pole which measure from the equator is 232.5 m/s.

Solution:
So the givens are, earth's radius = 6.37X10^6m, and the angular distance from the pole is 90 degrees. So 60 degrees is the 2/3.

r = 6.37x10^6 * cos(60) = 3.185x10^6m
since v = wr
v = 7.3x10^-5 * 3.185x10^6

v - 232.5 m/s
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Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

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     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

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Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

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substituting values

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   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

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Answer:

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