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topjm [15]
3 years ago
9

When it is burned, 1 gallon of gasoline produces 1.3Ã108j of energy. 1500 kg car accelerates from rest to 37 m/s in 10 s. the en

gine of this car is only 15% efficient (which is typical), meaning that only 15% of the energy from the combustion of the gasoline is used to accelerate the car. the rest goes into things like the internal kinetic energy of the engine parts as well as heating of the exhaust air and engine.\?
Physics
1 answer:
andreev551 [17]3 years ago
3 0
Im sorry I dont knoow this
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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
4 years ago
Imagine you are waiting for a train to pass at a railroad crossing. Will the train whistle have a higher pitch as the train appr
frosja888 [35]
THE DOPPLER EFFECT. Anyways, it would have a higher whistle as it approaches you, when it gets to you it only gets quieter because it leaves after. Think of a motorcycle going by, its loud coming to you then as it passes it gets quieter.
7 0
3 years ago
Materials that are soft and porous will absorb energy causing a decrease in amplitude and energy of the sound. This is called __
Vedmedyk [2.9K]
Damping of the sound
5 0
3 years ago
How long does it take for a Ford Econoline van moving at 39.5 m/s to travel 600 m?
Andrej [43]

Answer: B. 15.2s

Explanation: 600/39.5 = about 15.2

6 0
4 years ago
The security alarm on a parked car goes off and produces a frequency of 769 Hz. The speed of sound is 343 m/s. As you drive towa
uysha [10]

Answer:

Explanation:

ASSUMING your speed is constant

f₀ = f(v + vo)/(v + vs)

   Δf = f approach - f depart

69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

  vo = 15.5 m/s

6 0
2 years ago
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