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ehidna [41]
2 years ago
7

A satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther a

way from the same planet. How do the speeds of the two satellites compare
Physics
1 answer:
irinina [24]2 years ago
3 0

The speed of the second satellite is less than the speed of the first satellite.

<h3>What is speed?</h3>

The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.

Given is a satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther away from the same planet.

When the distance from the center of the orbit increases, the time to complete the orbit will be greater.

Thus, the speed of the second satellite is less than the speed of the first satellite.

Learn more about speed.

brainly.com/question/7359669

#SPJ1

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if we ever ride a airplane we dont mess up its signals and crash ,and its easier to ignore calls and texts

Explanation:

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Can you please help me? ​
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b= 26.2m

Explanation:

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Philosophy: The Big Picture Unit 8
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D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.
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A particle of mass m is placed in a three-dimensional rect- angular box with edge lengths 2L, L, and L. Inside the box the poten
QveST [7]

Answer:

the energy of groud state = \frac{9h^{2} }{32ml^{2} }

Explanation:

the energy of a 3 dimensional rectangular box is given by \frac{h^{2} }{8m}  (\frac{n_{x} ^{2} }{l_{x} ^{2}  }  +\frac{n_{y}^{2}  }{l_{y} ^{2} }+ \frac{n_{z}^{2}  }{l_{z} ^{2} })

where h is planks constant m is the mass of the particle n_{x},n_{y} and n_{z} are principal quantum number in x y and z direction. and l_{x},l_{y} and l_{z} are length of box in x y and z direction.

therefore the energy of ground state will be when n_{x},n_{y} and n_{z} = 1

therefore energy of ground state = \frac{h^{2} }{8m}  (\frac{1 ^{2} }{2l ^{2}  }  +\frac{1^{2}  }{l^{2} }+ \frac{1^{2}  }{l ^{2} })

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4 0
3 years ago
A closely wound rectangular coil of 100 turns has dimensions of 27.0 cm by 50.0 cm . The plane of the coil is rotated from a pos
kicyunya [14]

Answer:

\epsilon =71.677V induced.

Explanation:

We define all our variables,

B=1.2T

A=0.27m*0.50m

\theta= 90-37

\Delta T= 0.09s

N= 100

EMF induced is given through the expression

\epsilon = \frac{-N\Delta \Phi}{\Delta t}

Here we understand \Phi as

\Phi = BAcos\theta

We proceed to calculate the entire Initial Flow as the final, as well

\Phi_i=(1.2)(0.27*0.5)cos(90-37) = 0.09749Tm^2

Final Flow

\Phi_f=(1.2)(0.27*0.5)= 0.162Tm^2

Now, if \Delta t = 0.09s,

\epsilon= (100)(0.162Tm^2 - 0.09749Tm^2)/(0.09s)

\epsilon =71.677V induced.

NOTES:

  • <em>It is necessary to make two small notes regarding the development of the exercise. The subtraction of the angles is used since the exercise indicates that the angle is between the field B and the Plane. However, the measurement between the Area and the field is required b.</em>
  • <em>Negative signs can be neglected because it is understood that this is a reference to know which direction has the highest potential.</em>
8 0
4 years ago
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