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vichka [17]
3 years ago
6

The famous black planet, haunch, has a radius of 106 m, a gravitational acceleration at the surface of 4 m/s2 , and the tangenti

al speed of any point on its equator is 103 m/s. a famous haunch ant has a mass of 40 kg and is standing on a spring scale at the equator. what is the reading of the spring scale?
Physics
1 answer:
statuscvo [17]3 years ago
3 0
Gravity on the surface = 4 m/s^2
Now, the acceleration due to centripetal motion, a = v^2/R

Where,
v= 10^3 m/s, R = 10^6 m
Then,
a = (10^3)^2/(10^6) = 1 m^2/s

The net gravitational acceleration = 4-1 = 3 m/s^2

The reading on the spring scale = ma = 40*3 = 120 N
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What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass
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Answer:

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Explanation:

Given;

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let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz

The frequency of the simple pendulum is given as;

F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m

Thus, the length of the simple pendulum is 0.25 m.

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Climatology is a subspecialty of what?
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Explanation:

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As per ohm's law we will have

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