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vivado [14]
2 years ago
7

How does the work needed to lift an object and the gravitational potential energy of an object compare

Physics
1 answer:
Ganezh [65]2 years ago
8 0

When you do work to lift the object, the amount of work you do BECOMES the object's gravitational potential energy.  It GETS its potential energy from the work you do to lift it.  They're equal.  You lose it, and the object gains it.  Energy is not created or destroyed.  It's just transferred from you to the object.  

Later, when you DROP the object, GRAVITY does the same amount of work on it, to pull it to the ground.  Again, no energy is created or destroyed.  Every time a force acts to move anything, the energy to do it comes from somewhere, and the energy goes somewhere.

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6 0
3 years ago
For a satellite of mass mS in a circular orbit of radius rS around the Earth, determine its kinetic energy, K . Express your ans
agasfer [191]

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.

8 0
2 years ago
Question 2 10
ss7ja [257]
B Ik it all just know who they’ll you
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I feel like it could be A
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a 12000.0-kg car is traveling at 19 m/s. the driver suddenly slams on the brakes and skids to a stop. The coefficient of kinetic
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What I don’t get it ?
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