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Klio2033 [76]
3 years ago
13

Someone please help. Describe how electric potential energy, kinetic energy, and work change when two charges of opposite sign a

re placed near each other.
Physics
1 answer:
erik [133]3 years ago
5 0

Answer:

The answer is based on the conservation of energy law; something you should really understand by now.  

For convenience we can hold one of the two charges still; it becomes the frame of reference. And everything we say is in reference to the designated static charge, call it Q.  

So the moving charge, call it q, has total energy TE = PE. It's all potential energy as we start with q not moving.  

It has potential energy because in order to separate q from Q, we had to do work, add energy, on q. And from the COE law, that work added is converted into PE.  

It's a bit like lifting something off the ground. That's work and it becomes GPE. So there's some work, in separating the two charges in the first place.  

But there's more.  

Now we let q go. As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.  

And there's more work, done by the EMF on charge q. That converts the PE into KE and the q charge smashes into Q with some kinetic energy.

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A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.
WINSTONCH [101]

Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

Explanation:

Given that;

diameter of the mirror d = 1.7 m

height h = 180 km = 180 × 10³ m

wavelength λ = 500 nm = 5 × 10⁻⁹ m

Now Angular separation from the peak of the central maximum is expressed as;

sin∅= 1.22 λ / d

sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7

 sin∅ = 3.588 × 10⁻⁷

we know that;

 sin∅  = object separation / distance from telescope

object separation =   sin∅ × distance from telescope

object separation = 3.588 × 10⁻⁷  × 180 × 10³

object separation =6.45 × 10⁻² m

then we convert to centimeter

object separation = 6.45 cm

Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

5 0
3 years ago
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia
Eduardwww [97]

Answer:

From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km

So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr

So runner towards the west will be

distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8

So equating east and west time we have

x/9= (11-x)/8

8x=99-9x

17x=99

x=5.92 km

That is the distance covered by runner towards the east and he will meet the runner toward the west at

6-5.92=0.08 km west of the flagpole.

7 0
3 years ago
WILL GIVE BRAINLIEST PLZ HELP REALLY URGENT
ozzi
D.) 5kg

This is a trick question. The mass of an object does not change based in location. However the weight of an object does change, this is because Weight = Mass × Gravity. Also mass is measured in kilograms and so the answer is 5 kg. So if you ever want to lose weight just go to the moon!
3 0
3 years ago
A train travels 55 km south along a straight track in 34 minutes. What is the train's average velocity in kilometers per hour?​
AURORKA [14]
Hopefully I’m not late and I apologize if I am, but the answer to your question would be 95.6 km/hr. You know you can look up your question as well to see if they already have a answer to that so you won’t waste your points.
4 0
3 years ago
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Where is the electron located in an atom?
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Around the nucleus or in the shell of the atom. Hope this makes sense.
8 0
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