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LiRa [457]
3 years ago
5

An astronaut circling the earth at an altitude of 400 km is horrified to discover that a cloud of space debris is moving in the

exact same orbit as his spacecraft, but in the opposite direction. The astronaut detects the debris when it is 29 km away. How much time does he have to fire his rockets and change orbits?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
4 0

One of the essential concepts to solve this problem is the utilization of the equations of centripetal and gravitational force.

From them it will be possible to find the speed of the body with which the estimated time can be calculated through the kinematic equations of motion. At the same time for the calculation of this speed it is necessary to clarify that this will remain twice the ship, because as we know by relativity, when moving in the same magnitude but in the opposite direction, with respect to the ship the debris will be double speed.

By equilibrium the centrifugal force and the gravitational force are equal therefore

F_c = F_g

\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}

Where

m = mass spacecraft

v = velocity

G = Gravitational Universal Constant

M = Mass of earth

r \rightarrow R+h \Rightarrow Radius of earth and orbit

Re-arrange to find the velocity

\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}

\frac{v^2_{orbit}}{r} = \frac{GM}{r^2}

v^2_{orbit}=\frac{GM}{r}

v_{orbit} = \sqrt{\frac{GM}{r}}

v_{orbit} = \sqrt{\frac{GM}{R+h}}

Replacing with our values we have

v_{orbit} = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{6.37*10^6+0.4*10^6}}

v_{orbit} = 7676m/s

From the cinematic equations of motion we have to

t = \frac{d}{2v_{orbit}} \rightarrow Remember that the speed is double for the counter-direction of the trajectories.

Replacing

t = \frac{29000m}{7676m/s}

t = 3.778s

Therefore the time required is 3.778s

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The tension in the rope securing the upper pulley and the force that must be applied to keep the system in equilibrium ​
chubhunter [2.5K]

Answer:

Explanation:

In a frictionless system with no acceleration, the tension in the rope must be F along its entire length

FBD analysis of the lower pulley has two upward acting tension vectors F and one downward acting weight vector W

2F = W

F = W/2

FBD analysis of the upper pulley has one upward acting support vector T and three downward acting tension vectors F

T = 3F

T = 3(W/2)

T = 1.5W

8 0
3 years ago
A 0.0780 kg lemming runs off a
attashe74 [19]

Answer:

Ek = Ekv + Ekh = 4.101 + 0.914 = 5.015J

Explanation:

A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What is its potential energy (PE) when it lands

The potential energy PE, relative to the ground, will be zero, because the lemming is at the ground level.

HOWEVER, a much better question would be:

A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What is its kinetic energy (KE) when it lands?

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, u = 0, a = 9.81m/s^2, s = 5.36m

So we find v using equation (2)

v^2 = u^2 + 2as

v^2 = 0 + 2(9.81)(5.36) = 105.1632

So the kinetic energy resulting from the vertical drop is Ekv = ½mv^2

Ekv = ½(0.078)(105.16) = 4.101J

BUT we need to add in the kinetic energy resulting from the horizontal velocity, which did not change during the vertical drop.

Ekh = ½(0.078)(4.84^2) = 0.914J

So the total kinetic energy is Ek = Ekv + Ekh = 4.101 + 0.914 = 5.015J

5 0
3 years ago
You attend a football game one day and notice a long time difference between when the other side of the stadium stands up to che
viktelen [127]

Answer:if it were warmer i took the test i rly hope this helps ;)

Explanation:

6 0
3 years ago
In the theory of plate tectonics, various segments of Earths crust, called plates, move toward and away from each other. In one
GrogVix [38]

Answer:

-1.00 × 10⁻⁶ cm/y²

Explanation:

Using a = (v - u)/t where a = acceleration of the indian subcontinent, v = final speed of the indian subcontinent = 5 cm/y, u = initial speed of the indian subcontinent = 15 cm/y, and t = time of motion of the indian subcontinent = 1.00 × 10⁷ years.

Substituting the values of the variables, we find the acceleration of the indian subcontinent, a  = (v - u)/t = (5 cm/y - 15 cm/y)/1.00 × 10⁷ years = -10 cm/y ÷ 1.00 × 10⁷ years = -1.00 × 10⁻⁶ cm/y²

7 0
4 years ago
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aleksandrvk [35]
There are no options given in respect to the question and so it is not possible to choose. I would answer this question based on my knowledge and hope it satisfies you. he form of radio active decay that would be most likely detected by you if it were happening in the room next to the one you are currently standing in is gamma.
8 0
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