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ANTONII [103]
3 years ago
8

For a satellite of mass mS in a circular orbit of radius rS around the Earth, determine its kinetic energy, K . Express your ans

wer in terms of some or all of the variables mS, rS, and constants G and ME.
Physics
1 answer:
agasfer [191]3 years ago
8 0

Since it is asking you to find the kinetic energy in relation to the mass, radius, mechanical energy (total energy), and constants, you will need to setup an equation first to "find" the Mechanical Energy, so that we can then solve for the kinetic energy, as from my experience with high school physics, there is only the graviational potential energy equation and force in relation to celestial bodies.

Knowing the ME is the total energy, we add up the energies of the system. Since it is being influenced by the Earth, as per the problem stating the satellite has circular orbit around the Earth, we know there is gravitational potential. Since it is orbiting, we can assume some type of velocity. Nothing else that we need to worry about should be occuring at this level of physics, leaving you with

ME= Ug+K

from here we solve for K, as plugging in could get confusing and messy at the moment.

ME-Ug=K

now using the equations presumably given in class, if not then using this equation, we can find the Ug

Ug=(-(Gm*M)/r)     note that M is the mass of the Earth and m is the satellite

this should give us

ME-(-(GmM)/r)=K

since there is a negative being subracted, we can change that to

ME+(GmM)/r=K

I believe this should be fine, as the Earth's mass is constant, but if not, then all you need to figure left is how to get rid of the M in the equation, as the rest of the terms and constants are for sure within the requirements.

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I need help in my physics class and show me how it’s done
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If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)

Where | A | is the magnitude of the vector and \alpha is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle \alpha.

Vector A is in the 4th quadrant

So:

A_x = 6\\\\A_y = -6.5

So:

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So:

Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)

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Option 4.

4 0
3 years ago
What are the characteristics of high energy wave?
Ivahew [28]

Answer:

D. High frequency and short wavelengths.

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8 0
2 years ago
An emf of 22.0 mV is induced in a 519-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux
zhenek [66]

Answer:

\phi=1.56\times 10^{-5}\ Wb

Explanation:

Given that,

Emf, V = 22 mV

Number of turns in the coil us 519

Rate of change of current is 10 A/s.

We need to find the magnetic flux through each turn of the coil at an instant when the current is 3.70 A.

Let's find the inductance first. So,

L=\dfrac{\epsilon}{(dI/dt)}\\\\L=\dfrac{0.022}{10}\\\\L=0.0022\ H

We have,

L=\dfrac{N\phi}{I}, \phi is magnetic flux

\phi=\dfrac{LI}{N}\\\\\phi=\dfrac{0.0022\times3.7}{519}\\\\\phi=1.56\times 10^{-5}\ Wb

So, the magnetic flux is 1.56\times 10^{-5}\ Wb.

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3 years ago
A 873-kg (1930-lb) dragster, starting from rest completes a 401.4-m (0.2509-mile) run in 4.945 s. If the car had a constant acce
Delvig [45]

To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

x=x_0+v_0t+at^2

Where

x_0 = Initial position

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a = Acceleration

t= time

And the velocity can be expressed as,

v_f = v_0 + at

Where,

v_f = Final velocity

For our case we have that there is neither initial position nor initial velocity, then

x= at^2

With our values we have x = 401.4m, t=4.945s, rearranging to find a,

a=\frac{x}{t^2}

a = \frac{ 401.4}{4.945^2}

a = 16.41m/s^2

Therefore the final velocity would be

v_f = v_0 + at

v_f = 0 + (16.41)(4.945)

v_f = 81.14m/s

Therefore the final velocity is 81.14m/s

8 0
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