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pochemuha
3 years ago
15

If a refrigerator is a heat pump that follows the first law of thermodynamics, how much heat was removed from food inside of the

refrigerator if it released 380J of energy to the room?
Physics
2 answers:
Anna007 [38]3 years ago
7 0
Idk I haven’t learned about this yet
Katena32 [7]3 years ago
7 0

A refrigerator considered as a heat pump, releases 380 J of energy into the room. 380 J of energy was removed from the food.

<u>Explanation:</u>

The first law of thermodynamics is related to the law of conservation energy which states that energy can be  transferred from one form to another but can be neither created nor destroyed.

If 380 J of energy was released into the room by refrigerator, then 380 J of energy was removed from the food inside the refrigerator.

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El calor es la transferencia de energía de un objeto más cálido a un objeto más fresco. El calor puede transferirse de tres maneras: por conducción, por convección y por radiación. La conducción es la transferencia de energía de una molécula a otra por contacto directo.
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3 years ago
What is emotional strength?​
Ugo [173]

Answer:

Emotional strength is a person's ability to deal with challenges (any kind) and how they can bounce back on that certain situation.

I hope this helps you.

4 0
3 years ago
Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m
Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5  *10⁻⁹C

r₁=0.840m

r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}

sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246

cos\beta =\frac{1}{\sqrt{1.64} } =0.7808

Calculation of the electric field at point P due to q1

Ep₁x=0

Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2}  } =\frac{8.99*10^{9}*2.9*10^{-9}  }{0.84^{2} } =36.95\frac{N}{C}

Calculation of the electric field at point P due to q2

Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2}  } =-21.4\frac{N}{C}

Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2}  } =-17.11\frac{N}{C}

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0
3 years ago
Is the voltage of two identical lamps the same?​
Dahasolnce [82]

Answer:

It depends if they have the same lightbulb in them.

Explanation:

8 0
2 years ago
Read 2 more answers
I need it in the next hour or so!
PSYCHO15rus [73]

The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is

∑ F = F[a] - F[f] - F[air] = ma

3100 N - 200 N - F[air] = (650 kg) (3 m/s²)

Solve for F[air] :

F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)

F[air] = 3100 N - 200 N - 1950 N

F[air] = 950 N

3 0
2 years ago
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