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Rasek [7]
3 years ago
15

Consider a satellite in a circular orbit around the earth. Why is it important to give a satellite a horizontal speed when placi

ng it in orbit? What will happen if the horizontal speed is too small? What will happen if the speed is too large?
Physics
1 answer:
Andrei [34K]3 years ago
4 0

Answer:

In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,

Explanation:

When a satellite is in orbit the most important force is the docking of gravity with the Earth

             F = m a

where the acceleration is centripetal and F is the force of universal attraction

centripetal acceleration is

             a = v² / r

             

             F = m v² / r

In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.

The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.

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Suppose Earth's mass increased but Earth's diame-
navik [9.2K]

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

F = G \frac{m1m2}{r^2}

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

7 0
3 years ago
The mass of the boy is 35 kilograms. What is the force of friction that slowed him down?
pychu [463]

Answer:

Newton's F=ma, which means the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a)

3 0
3 years ago
Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441
Tasya [4]
The period of a simple pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
g= \frac{4 \pi^2}{T^2}L (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz
And its period is the reciprocal of its frequency:
T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s

So now we can use eq.(1) to find the gravitational acceleration of the planet:
g= \frac{4 \pi^2}{T^2}L =  \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2
3 0
3 years ago
A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion in
Rashid [163]

Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J  

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J  

But  ,

4 * 1875000 = 7500000

so the KE has increased by 4 times.

8 0
3 years ago
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NEED HELP NOW ETHANHUNT 25 POINTS WILL MARK BRAINLIEST!! For the independent reading all you have to do is pick a grade six book
garik1379 [7]
Can you please stop pasting this question, just go to his profile and ask him.
7 0
3 years ago
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