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harina [27]
3 years ago
12

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a

satellite in an orbit 900 km above Earth’s surface. (b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite’s orbit at an angle of 90º relative to Earth. What is the velocity of the rivet relative to the satellite just before striking it?
Physics
1 answer:
Pie3 years ago
6 0

Answer:

Part a)

v = 7407.1 m/s

Part b)

v_{rel} = 1.05 \times 10^4 m/s

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

v = \sqrt{\frac{GM}{R+h}}

here we know that

M = 5.98 \times 10^{24} kg

R = 6.37 \times 10^6 m

h = 900 km = 9.0 \times 10^5 m

now we have

v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}

v = 7407.1 m/s

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

v_{rel} = \sqrt{2} v

v_{rel} = 1.05 \times 10^4 m/s

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Vesnalui [34]

Answer:

C. 2.5 N

Explanation:

The sum of downwards moments equals sum of moments upwards

<u>Downwards moments </u>

Moments= Force*distance

               = [10/100 * 2 ] +[20/100*4]

               =[0.1*2] +[0.2*4]

               =[0.2+0.8]

               = 1.0 N

Equate this answer to upward force ;

1.0 N = 0.4 *U

1/0.4 =U

2.5 N =U

8 0
3 years ago
Visually, how can you distinguish between an ac and dc power supply?
Temka [501]

Answer:

By the use of slow motion camera.

Explanation:

Visually, it is very hard to differentiate between an ac and dc power supply. But Since, we that In Ac supply polarity changes 100 times in a second ( because frequency of ac supply is 50 Hz generally). Whereas, Dc gives a steady power supply. So, in slow motion camera we can easily capture the flickering light tubes which won't happen in case of dc supply.

5 0
3 years ago
Read 2 more answers
A flywheel with a very low friction bearing takes 1.6 h to stop after the motor power is turned off. The flywheel was originally
nadezda [96]

Answer: 5.76 rads/s

Explanation:

The initial rotation is 55 rpm

1 rev = 2π radians

55 revs = 55 × 2π/1 = 345.58 radians/min

345.58 rads/min = 345.58rads/60s = 5.76 rads/s

3 0
4 years ago
This is physics 11th grade and a homework question I don’t understand how to do this or what the question is asking me
Alexxx [7]

a) Frequency is the number of complete oscillations per second. Looking at the graph, there are 9 complete oscillations in 5 seconds. Thus,

Frequency = 9/5 = 1.8 oscillations per second

Frequency = 1.8 Hz

Period = 1/frequency = 1/1.8

Period = 0.056 s

b) When we differenctiate displacement with respect to time, the result is velocity.

Recall, period = 1/f = 5/9 cycles

1/4 cycle behind = 1/4 x 5/9 = 5/36

It is delayed with 5/36 sec with respect to displacement.

5/36 sec = 0.139 sec

Acceleration = first derivative of velocity = second derivative of displacement = 1/4 cycle behind velocity = 1/2 cycle behind displacement =

5/36 = 0.139 sec delayed with respect to velocity

= 5/18 = 0.2777 secs delayed with respect to displacement

Thus, the number of seconds out of phase with the displacements is 0.278 seconds

c) The formula for calculating the period of an ideal pendulum anywhere is

T = 2π√length/local gravity). We would calculate the local gravity.

From the information given,

length = 0.2

T = P = 5/9

Thus,

5/9 = 2π√0.2/local gravity)

(5/9)/2π = √0.2/local gravity

Square both sides. It becomes

[(5/9)/2π]^2 = 0.2/local gravity

local gravity = 0.2/[(5/9)/2π]^2

local gravity = 25.56 m/s^2

Thus,

acceleration due to gravity = 25.56 m/s^2

Recall, earth's gravity = 9.8 m/s^2

number of g forces = 25.56/9.8

number of g forces = 2.61

6 0
1 year ago
A ball is rolled uphill a distance of 5 meters before it slows, stops, and begins to roll back. The ball rolls downhill 9 meters
Zepler [3.9K]
The magnitude of the ball's displacement when the ball is rolled uphill with a 5 meters before it slows, stops and begins to roll back and the ball rolls downhill 9 meters before coming to rest is 4 meter. Just subtract 9 meters with 5 meters. Therefore the answer is 4 meters. 
6 0
3 years ago
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