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olya-2409 [2.1K]
4 years ago
8

A satellite in orbit is not truly traveling through a vacuum.It is moving through very, very thin air. Does the resulting airfri

ction cause the satellite to slow down?
b) Explain the similarities anddifferences between Newton’s law of universal gravitation andCoulomb’slaw.
Physics
1 answer:
defon4 years ago
4 0

PART A) Yes, the fact that there is a frictional force acting on the satellite generates a loss of energy due to friction. What causes satellite to diminish its orbit during its tour. In fact, many satellites have rectifier systems that allow them to position themselves and remain in their orbit for a long time to avoid being trapped by the Earth's gravity Force and fall into the atmosphere where they would probably be torn apart.

PART B) As a similarity, one could start by mentioning the structure of the two equations are similar and have their own constants who were responsible for supporting them. While the law of gravity speaks of the masses of the bodies the electrostatic law speaks of the charges of the bodies. For both the force is inversely proportional to the square of the distance that separates them.

However, the most notable difference between them is basically their statement. While one of the equations speaks about greavedad the other reflects the electromagnetic phenomena. It should be noted that the force of gravity is much weaker than the electromagnetic force and that the latter has the capacity of attraction and repulsion. While the gravitational force only that of attraction.

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A person takes a trip, driving with a constant speed of 89.5km/h, except for a 22.0−min rest stop. If the person's average speed
Elanso [62]

Answer:

stop using brainly and learn this bs in class

Explanation:

8 0
4 years ago
After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m long and 7.8 mm in diameter
Nesterboy [21]

Explanation:

Given that,

Mass of the rock climber, m = 90 kg

Original length of the rock, L = 16 m

Diameter of the rope, d = 7.8 mm

Stretched length of the rope, \Delta L=3.1\ cm

(a) The change in length per unit original length is called strain. So,

\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193

(b) The force acting per unit area is called stress.

\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa

(c) The ratio of stress to the strain is called Young's modulus. So,

Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2

Hence, this is the required solution.

8 0
3 years ago
Diamonds are a very dense material. Predict what would happen to the light ray if you projected it from air through a diamond
Kipish [7]

Answer:

It [the diamond] would act like a prism, and make a rainbow, or, the light would break up and disappear

Explanation:

that's what I would think at least

4 0
2 years ago
Read 2 more answers
In 2 minutes, a ski lift raises four skiers at a constant speed to a height of 123 m. The average mass of each skier is 71 kg. W
serious [3.7K]

Answer:

The average power  provided by the tension in the cable pulling the lift is = 714 W

Explanation:

Given data

Mass = 71 kg

Change in height = 123 m

When the lift moves in upward direction then in that case kinetic energy is constant & only potential energy changes.

Change in potential energy Δ PE = m g ( h_f - h _ i )

Δ PE = 71 × 9.81 × 123

Δ PE = 85670.73 J

Time = 2 min = 120 sec

So average power is given by

P_{avg} = \frac{Change \ in \ potential \ energy}{Time}

P_{avg} = \frac{85670.73}{120}

P_{avg} = 714 \ W

Therefore the average power  provided by the tension in the cable pulling the lift is = 714 W

7 0
3 years ago
calculate the eccentricity of jupiter's orbit if the aphelion is 816.62 million km and the perihelion is 741.52 million km
schepotkina [342]
<h2>Answer:0.048</h2>

Explanation:

Let the length of perihelion be p

Let the length of aphelion be a

Let the length of semi major axis of the elliptical path be s

As we know that semi major axis is the average of perihelion and aphelion.

So,s=\frac{p+a}{2} =\frac{1558.14}{2}=779.07millionKm

Let the eccentricity of the elliptical path be e

e=1-\frac{p}{s}=1-\frac{741.52}{779.07}=0.048

8 0
3 years ago
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