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mrs_skeptik [129]
2 years ago
7

Does Drag Oppose Lift​

Physics
1 answer:
katrin2010 [14]2 years ago
8 0
I do not understand your question..
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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
A steel cable has a cross-sectional area 4.49 × 10^-3 m^2 and is kept under a tension of 2.96 × 10^4 N. The density of steel is
Lemur [1.5K]

Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

Explanation:

let T = 2.96×10^4 N be the tension in in the steel cable, ρ  = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

ρ = m/V

m = ρ×V

but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

m/L = ρ×A

then the speed of the wave in the cable is given by:

v = √(T×L/m)

  = √(T/A×ρ)

  = √[2.96×10^4/(4.49×10^-3×7860)]

  = 25.5 m/s

Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

7 0
3 years ago
When driving at night, only use your high-beam headlights___
zzz [600]

Answer:

B

Explanation:

While answer C may sound correct, Answer B is makes more sense. We know you cant use High-beam lights when u cant see ongoing traffic because it could affect the other driver coming across from you. Its good to use it when legal and safe, but in that term I still don't believe there's no reason for HIGH-beamed. That's this leaves B, when you are on u lighted streets.

6 0
2 years ago
The charges that are free to move in a metallic conducting wire and that are responsible for the flow of electric current are- a
SVEN [57.7K]

Answer:

D) The negatively charged electrons

Electricity passes through metallic conductors as a flow of negatively charged electrons. The electrons are free to move from one atom to another. We call them a sea of delocalised electrons. Current was originally defined as the flow of charges from positive to negative. Please give me the brainliest answer?

:) Hoped this helped!!! Have a good day!!! <3

3 0
2 years ago
The periodic table was developed in order to group and organize elements with similar physical and chemical properties.
nadezda [96]
Your answer to this question is A halogens. 
6 0
3 years ago
Read 2 more answers
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