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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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2 years ago
A pilot, whose mass is 84.0 kg, makes a loop-the-loop in a fast jet. Assume that the jet maintains a constant speed of 345 m/s a
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Answer:

The apparent weight is 5 times greater than the original weight at the bottom.

Explanation:

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Mass of the pilot, m = 84 kg

Velocity of the jet, v = 345 m/s

Radius of the loop, R = 3.033 km = 3.033 * 10^3 m

We have to find the apparent weight that the pilot feels.

Let the apparent weight be "N" .

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From the FBD shown we can say that :

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⇒ N=mg(1+\frac{v^2}{Rg} )

⇒ N=mg(1+\frac{(345)^2}{3.033\times 10^3 \times 9.8} )

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