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Lorico [155]
3 years ago
15

ANSWER One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is

at rest as it hangs from the string. When the object is raised to a height above its lowest point and released from rest, the object undergoes simple harmonic motion with a frequency f0. In a second scenario, the length of the string is cut in half before the object undergoes simple harmonic motion again. What is the new frequency of oscillation of the object in terms of f0?
Physics
1 answer:
qaws [65]3 years ago
4 0

Answer:

Explanation:

It is a case of oscillation by simple pendulum . Expression for simple pendulum is given as follows

T = 2\pi\sqrt{\frac{l}{g} }

where T is time period , l is length of pendulum and g is acceleration due to gravity .

\frac{1}{f} =2\pi\sqrt{\frac{l}{g} }  , f is frequency of oscillation

For the given case

\frac{1}{f_o} =2\pi\sqrt{\frac{l}{g} }

subsequently length becomes half so

\frac{1}{f} =2\pi\sqrt{\frac{l}{2g} }

dividing

\frac{f}{f_o} = \sqrt{\frac{2}{1} }

f = \sqrt{2} f_o

frequency of oscillation becomes √2 times.

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Answer:

the car have travelled 0.31 mile during that​ time

Explanation:

Applying the Equation of motion;

s = 0.5(u+v)t

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u = initial speed = 0 mph

v = Final speed = 50 mph

t = time taken = 3/4 min = 3/4 ÷ 60 hours = 1/80 hour

Substituting the given values into the equation;

s = 0.5(0+50)×(1/80)

s = 0.3125 miles

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An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr
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An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

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