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Bas_tet [7]
3 years ago
7

An object 1 of mass m1 is separated by some distance d from an object 2 of mass 2m1 . An object 3 of mass m3 is to be placed bet

ween them. If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?
Physics
1 answer:
Harrizon [31]3 years ago
6 0

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

 

Object 3 should be placed closer to object 1.

 

Object 3 should be placed on a halfway between object 2 and object 1.

 

Object 3 should be placed closer to object 2.

 

Solution

I think that Object 3 should be placed closer to object 2.

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it's important because it shows how thermal energy transforms or continues to be all around us in everything

6 0
2 years ago
Pete is driving down 7th Street. He drives 300 meters in 18 seconds. Assuming he does not speed up or slow down, what is his spe
Korolek [52]

Answer:

16.67m/s

Explanation:

Given parameters:

Distance Pete drove  = 300m

Time taken  = 18s

Unknown:

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Solution:

Speed is the distance traveled per unit of time.

It is mathematically expressed as;

   Speed  = \frac{distance}{time}

Insert the parameters and solve;

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3 0
2 years ago
A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d
lisabon 2012 [21]

The centripetal acceleration a is 4.32 \times 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

Given, distance = 400 mm = 0.4 m,       Time = 100 s.

Computing the v = 0.4 m / 100 s

                         v = 4 \times 10^-3 m/s.

radius of the circular end r = 37 mm = 0.037 m.

            centripetal acceleration a = v^2 / r

                                                        = (4 \times 10^-3)^2 / 0.037

                                                    a = 4.32 \times 10^-4 m/s^2.

6 0
2 years ago
Suppose that two objects attract each other with a gravitational force of 32 Newtons. If the distance between the two objects is
Crazy boy [7]

Answer:

64N

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Have a wonderful day!
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3 years ago
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