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liberstina [14]
2 years ago
8

Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellit

e X has an orbital radius of 3R, and satellite Y has an orbital radius of 4R. The kinetic energy of satellite X is Kx.
1) In terms of Kx, the gravitational potential energy of the planet-satellite X system is
(A) -2Kx
(B) -Kx
(C) -Kx/2
(D) Kx/2
(E) 2Kx

2) Satellite x is moved to the same orbit as satellite Y by a force doing work on the satellite. In terms of Kx, the work done on satellite X by the force is
(A) -Kx/2
(B) -Kx/4
(C) 0
(D) +Kx/4
(E) +Kx/2

Show the work for each question.
Physics
1 answer:
GrogVix [38]2 years ago
3 0

Answer:

1) C

2) E

Explanation:

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If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed t
Whitepunk [10]
From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)

Distance traveled is
d = 100 km = 10⁵ m

Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s

Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s

The difference in time is
302.12 - 19.49 = 282.63 s

Answer:  283 s (nearest second)



6 0
2 years ago
A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp
liq [111]

Answer:

W_2=\sqrt{\frac{3}{5} }W_1

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

I_1 =\frac{2}{5}MR^2

K_1 = \frac{1}{2}IW_1^2

So, replacing, we get that:

K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

At the same way, the moment of inertia and kinetic energy for second ball is:

I_2 =\frac{2}{3}MR^2

K_2 = \frac{1}{2}IW_2^2

So:

K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2

Then, K_2 is equal to K_1, so:

K_2 = K_1

\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2

\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2

Finally, solving for W_2, we get:

W_2=\sqrt{\frac{3}{5} }W_1

5 0
2 years ago
Discuss the production, transmission, and usage of electricity in the context of conservation of energy. When electricity is “us
mr Goodwill [35]

There are huge losses in the transmission, production and usage of electricity and the reduction of these losses in order to save electricity is called as conservation of energy.

As per the statistics, there is loss of nearly 4% while the transmission of electricity. Like wise during production also, lot of electricity get wasted due to the inefficient material used. None of the production material nor the equipment used have 100% efficiency and thus there is always a possibility of energy wastage.

When it is said that the energy is wasted , it simply means that the energy production which should have been 100% as per calculation is not completely derived from the source due to the inefficient conversion process. For example, a turbine while rotating must convert 100 % of the water energy or water falling on it into electrical energy but the turbine is not able to do so as some of the water is lost or its energy is lost before conversion while going through the mechanical process.

3 0
2 years ago
The potential energy of a 40 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?
Fynjy0 [20]
The formula for potential energy is PE=mgh
It can have that high of a potential energy if it's relative height what super high.
7 0
2 years ago
An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t
marissa [1.9K]
The energy transferred to the spring is given by:
U= \frac{1}{2}kx^2
where 
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x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
k=52.9 N/cm = 5290 N/m
x=40.0 cm=0.4 m

so now we can use these data inside the equation ,to find the energy transferred to the spring:
U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J
4 0
2 years ago
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