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solong [7]
3 years ago
14

The force of gravity considers which two parts?

Physics
1 answer:
attashe74 [19]3 years ago
3 0
Mass of object/source
Gravity
Fg=mg
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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
mestny [16]

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.

3 0
3 years ago
Llus
ivann1987 [24]

Answer:

kwkwskdkmfmfmdkakksjddndkkakamsjjfjdmakisnddnianbfdnnskake

now

8 0
3 years ago
Read 2 more answers
what output force is generated when an input force of 630 n is applied to a machine with a mechanical advantage of 3
Marta_Voda [28]
The mechanical advantage is the factor by which
the machine multiplies the input force.

If the MA is 3 and the input force is 630N, then
the output force is

           (3) x (630N) = 1,890N
3 0
3 years ago
Read 2 more answers
Leonard is stressed and feeling overwhelmed by the college selection process. What should he do to combat this problem?
exis [7]
He should ask other people for advice
6 0
3 years ago
A person's prescription for her new bifocal glasses calls for a refractive power of -0.450 diopters in the distance-vision part,
Angelina_Jolie [31]

Answer:

Far point of the eye is 22.24 m

Far point of the eye is 0.4 m

Explanation:

\frac{1}{f}=-0.045

Object distance = u

Image distance = v

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=-0.045-\frac{1}{\infty}\\\Rightarrow \frac{1}{v}=\frac{1}{-0.045}\\\Rightarrow v=-22.22\ m

Far point

|v|+\text{Position from eye}\\ =|-22.22|+0.02\\ =22.24\ m

Far point of the eye is 22.24 m

Object distance = u = 0.25-0.02 = 0.23 m

\frac{1}{f}=1.75

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=1.75-\frac{1}{0.23}\\\Rightarrow v=-0.38\ m

Near point

|v|+\text{Position from eye}\\= |-0.38|+0.02\\ =0.4\ m

Far point of the eye is 0.4 m

7 0
3 years ago
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