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NARA [144]
2 years ago
11

A system initially has an internal energy U of 504 J. It undergoes a process during which it releases 111 J of heat energy to th

e surroundings, and does work of 222 J. What is the final energy of the system, in J?
Physics
1 answer:
Likurg_2 [28]2 years ago
6 0

Answer:

615 J

Explanation:

internal energy (U) = 504 J

heat lost (q) = 111 J = - 111 J (negative sign is because heat is lost)

work done = 222 J

what is the final energy in the system

total energy = final energy - initial energy

final energy = total energy + initial energy

where

initial energy = 504 J

total energy = 222 - 111 = 111 J

final energy = 504 + 111 = 615 J

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If the sun suddenly ceased to shine, how long would it take earth to become dark? you will have to look up the speed of light in
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We are 8 light minutes from the sun. That means two things, we see the sun as it was 8 minutes ago, and we WOULD continue to see the sun for 8 minutes after it disappeared.
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3 years ago
How are some types of collisions different from others?
lyudmila [28]
There are two general types of collisions, inelastic and elastic. 
Inelastic collisions occur when two objects collide but neither of them bounce away from each other.
Collisions in which the objects do not touch each other are elastic. (Ex: Rutherford Scattering) 
3 0
2 years ago
You add 800 ml of water at 20c to 800 ml of water at 80c what is the most likely final temperature of the mixture ?
bekas [8.4K]

Answer:

d. 50 C

Explanation:

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According to the 2nd law of thermodynamics, heat transfers from hot to cold temperature.

The quantity of both the different waters is equal so this makes it very easy. All we have to do is find the mean of both the temperatures:

Final temperature = (20 C + 80 C)/2

= 50 Celsius

3 0
3 years ago
Read 2 more answers
7. How much work is done in moving a charge of 10 micro coulombs 1 meter along an equipotential of 10 volts?
Neko [114]

It takes work to push charge through a change of potential. 
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so that path doesn't require any work.

8 0
3 years ago
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three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25
ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0
2 years ago
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