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nekit [7.7K]
3 years ago
5

If the vertical axis and time on the horizontal axis if the speed is steadily increasing could the speef line ever become perfec

tly vertical
Physics
1 answer:
Phantasy [73]3 years ago
8 0

A graph of real speed can have a section that's as steep as you want,
but it can never be a perfectly vertical section.

Any vertical line on a graph, even it it's only a tiny tiny section, means
that at that moment in time, the speed had many different values.

It also means that the speed took no time to change from one value to
another, and THAT would mean infinite acceleration.

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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
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
3 years ago
If a train is travelling 200km/hour eastward for 1800 seconds how far does it travel?
Olenka [21]

Answer:

Distance, d = 99990 meters

Explanation:

It is given that,

Speed of the train, v = 200 km/h = 55.55 m/s

Time taken, t = 1800 s

Let d is the distance covered by the train. We know that the speed of an object is given by total distance covered divided by total time taken. Mathematically, it is given by :

v=\dfrac{d}{t}

d=v\times t

d=55.55\times 1800

d = 99990 m

So, the distance covered by the train is 99990 meters. Hence, this is the required solution.

5 0
3 years ago
How much heat is given off when 210.0g of water at 0 degrees freezes into ice?
Sergio [31]

In the freezing physical change, when 210.0 g of water a 0 degrees freezes into ice, it gives off 71.0 kJ of heat.

<h3>What is freezing?</h3>

It is a physical change in which liquids give off heat to form solids.

We have 210.0 g of water at 0°C. We can calculate the amount of heat given off when it freezes into ice using the following expression.

Q = ΔH°fus × m

Q = 0.334 kJ/g × 210.0 g = 70.1 kJ

where,

  • Q is the heat released.
  • ΔH°fus is the latent heat of fusion.
  • m is the mass.

In the freezing physical change, when 210.0 g of water a 0 degrees freezes into ice, it gives off 71.0 kJ of heat.

Learn more about freezing here: brainly.com/question/40140

#SPJ1

5 0
2 years ago
The change of motion of an object depends on ___
Gennadij [26K]
A is the answer mass and force
3 0
3 years ago
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