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aleksandrvk [35]
3 years ago
6

Tidal Forces near a Black Hole. 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.
(a) What is the tension between her ears?
Physics
1 answer:
arlik [135]3 years ago
5 0

Answer:

833.4801043*10^6N on ear that is closer to Black hole.

13.83803929*10^6N On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

                                                 F =\frac{m_{1}*m_{2}  }{r^2} G

                    m_{1}= 5*1.989*10^30kg is mass of Black hole.

                     m_{2}= 0.030kg is mass of ear that is close to black hole.

                    G = 6.679*10^-11m^3*kg^-1*s^-2

                     r = 120km-\frac{6}{100000} km=119.99994km=119999.94m

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

                                      F = 833.4801043*10^6N

       That is tension of ear.

(2) Force on ear that is further from black hole.

                              F =\frac{m_{1}*m_{2}  }{r^2} G

                    m_{1}= 5*1.989*10^30kg is mass of Black hole.

                     m_{2}= 0.030kg is mass of ear that is close to black hole.

                    G = 6.679*10^-11m^3*kg^-1*s^-2

         this time r is further away from black hole so it would be.

                       r = 120km+\frac{6}{100000} km = 120000.06m

Plugging this all in we get

                          F = 13.83803929N

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.

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Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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How much power is needed to lift a 20-kg object to a height of 4.0 m in 2.5 seconds?
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Answer:

313.92w

Explanation:

Formula for power:

P=W/∆t = Fv

Givens:

m=20kg

∆y=4.0m

∆t=2.5s

a=9.81m/s²

In order to find power, we first need to solve for work.

W=Fd (force*displacement), f=mg

W=mg∆y

W=(20kg)(9.81m/s²)(4.0m)

W=784.8J

P=W/∆t

P=784.8J/2.5s

P=313.92 watts

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Students perform an experiment in which they drop two eggs with equal mass from a balcony. In the first trial, the egg hits the
shepuryov [24]

<u> Answer </u>

The impulse on the second trial is smaller is smaller than in the first trial.

<u>Explanation </u>

Impose of a body is that change in momentum during a time interval. If the change of momentum takes longer then, the impulse of a force is less. I a moving object hits a hard surface the rate of change of momentum is very high. e.i in the first trial, the egg breaks because it hits the hard surface(ground).

In the second trial, the foam cushion absorbs the shock and prolongs the time of impact with the egg hence decreasing the impulse.


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A cyclist going downhill is accelerating at 1. 2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is
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Answer:

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Explanation:

We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.

v_f= v_i + at

The cyclist is acceleration at 1.2 meters per second squared. After 10 seconds, the velocity is 16 meters per second.

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Substitute the values into the formula.

16 \ m/s = v_i + (1.2 \ m/s^2)(10 \ s)

Multiply.

16 \ m/s = v_i + (1.2 \ m/s^2 * 10 \ s)

16 \ m/s = v_i + 12 \ m/s

We are solving for the initial velocity, so we must isolate the variable v_i. Subtract 12 meters per second from both sides of the equation.

16 \ m/s - 12 \ m/s = v_i + 12 \ m/s -12 \ m/s

4 \ m/s = v_i

The cyclist's initial velocity is <u>4 meters per second.</u>

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