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Inessa [10]
4 years ago
12

The independent variable:

Physics
1 answer:
ValentinkaMS [17]4 years ago
3 0

Answer:

A. Is the one that the experimenter manipulates directly

Explanation:

The independent variable is the one that is manipulated during an experiment by the experimenter.

The dependent variable is the one that is effected by the independent variable in an experiment.

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The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at C ,
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2 years ago
Determine the rate at which the electric field changes between the round plates of a capacitor, 5.8 cm in diameter, if the plate
Natalka [10]

Answer:

The rate at which the electric field changes between the round plates of a capacitor is 125\times 10^{3}Vs^{-1}.

Explanation:

It is given in the problem that the round plates of a capacitor are spaced some distance apart and the voltage across them is changing.

The expression for the electric field in terms of voltage is as follows;

E=\frac{V}{d}

Here, E is the electric field, V is the voltage and d is the distance of separation.

Differentiate expression of the electric field with respect to time, t.

\frac{dE}{dt}=\frac{1}{d}\frac{dV}{dt}

Convert the distance of separation from mm to m.

d= 1.2 mm

d=1.2\times 10^{-3}m

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\frac{dE}{dt}=\frac{1}{d}\frac{dV}{dt}

Put \frac{dV}{dt}=150 Vs^{-1} and d=1.2\times 10^{-3}m

\frac{dE}{dt}=\frac{1}{1.2\times 10^{-3}}(150)

\frac{dE}{dt}=125\times 10^{3}Vs^{-1}

Therefore, the rate at which the electric field changes is 125\times 10^{3}Vs^{-1}.

4 0
3 years ago
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