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MArishka [77]
2 years ago
11

PLEASE HELP!!!!

Physics
1 answer:
Elena L [17]2 years ago
6 0

The emf power supply should be used for a 0.75 A recharging current is 2.25 V.

<h3>What is current?</h3>

The current is the stream of charges which flow inside the conductors when connected across the end of voltage.

Given is the battery having 1.5 V open circuit potential and internal resistance of 1 ohm.

From the Ohm's law, V =IR

R = V/I

Here, R is the proportionality constant.

To charge current I = 0.75 A

Voltage V = 0.75A x 1 ohm = 0.75 V

Total potential required is

= 1.5 +0.75

= 2.25 V

Thus, the emf power supply should be used for a 0.75 A recharging current is 2.25V.

Learn more about current.

brainly.com/question/10677063

#SPJ1

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An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency o
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Answer:

Part A

The efficiency of the engine is 10%

Part B

The change in internal energy is 300 J

Part C

The change in volume is 1 m³ which is one cubic meter

Explanation:

Part A

Efficiency is defined as the ratio of energy output to energy input;

The given parameters are;

The distance the car is moved, d = 10 meters

The force which moves the car, F = 100 N

The amount of energy produced by the engine = 1,000 J

Therefore, we have;

The energy output to the car = The work done on the car = Force applied to the car, F × The distance the car moves, d;

∴ The energy output to the car by the engine = F × d = 100 N × 10 m = 1,000 J

The energy input from the engine = The energy produced by the engine = 10,000 J

The efficiency of the engine = (The energy output)/(The energy input)= 1,000J/10,00J = 0.1

The efficiency in percentage = 0.1 × 100 = 10 %

The efficiency of the engine = 10%

Part B

The amount of heat added to the substance, ΔQ = 1,000J

The amount of work the substance does on the atmosphere, W = 700 J

The change in internal energy, ΔU is given as follows;

ΔQ = W + ΔU

∴ ΔU = ΔQ - W

For the substance, we have;

The change in internal energy, ΔU = 1,000 J - 700 J = 300 J

Part C

The work done by the piston, W = 1,000 J

The pressure, in the piston, P = 1,000 Pa = constant

The work done by the piston in a constant pressure process, W = P × ΔV

Where;

W = The work done

P = The constant pressure applied

ΔV = The change in volume = V₂ - V₁

V₂ = The final volume

V₁ = The initial volume

∴The change in volume ΔV = W/P = 1,000 J/(1,000Pa) = 1 m³

The change in volume ΔV = 1 m³

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