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never [62]
3 years ago
7

Diabatic processes occur when:

Physics
1 answer:
Svetach [21]3 years ago
4 0

Answer:

The correct answer is d. add or remove energy from a system.

Explanation:

A diabatic process can be defined as one where a system exchanges energy with the surroundings resulting in a thermodynamic change of the system's state.

Diabatic processes can be grouped into five categories viz

1. Terrestrial radiation.

2. Solar radiation

3. latent heat flux

4. sensible heat flux

5. friction process

In the same vein, an adiabatic process is one in which there is zero heat, mass or momentum transfer across the bounary of the thermodynamic  system

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fomenos

Answer:

protons

Explanation:

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3 years ago
Define 1 unit electricity​
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Answer:

A unit is represented in kWH or Kilowatt Hour. This is the actual electricity or energy used. If you use 1000 Watts or 1 Kilowatt of power for 1 hour then you consume 1 unit or 1 Kilowatt-Hour (kWh) of electricity.

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Consider an unknown charge that is released from rest at a particular location in an electric field so that it has some initial
sineoko [7]

Answer:

b)

Explanation:

If the charge is released at rest in an electric field, it will move along the electric field, going to regions of higher electric potential if it is a negative charge (against the field direction) and towards lower potential regions if it is positive (along the field). This means that the charge will gain kinetic energy, energy that only can come from a decrease in the electric potential energy.

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For a negative charge: ΔEp = (-q) *ΔV < 0 (as ΔV > 0)

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In 1923, the United States Army (there was no U.S. Air Force at that time) set a record for in-flight refueling of airplanes. Us
dybincka [34]

Answer:

1.95m/s

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Please view the attached file for the detailed solution.

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Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan
Drupady [299]

Answer:

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

Explanation:

We are given that

Gravitational force=F_g=\frac{Gm_Emr}{R^3_E}

r=0,U(0)=0

We know that

Gravitational potential energy=-\int F_gdr

U(r)=-\int\frac{Gm_Emr}{R^3_E}dr

U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C

Substitute r=0 ,U(0)=0

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U(r)=-\frac{Gm_Emr^2}{2R^3_E}

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