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olchik [2.2K]
3 years ago
5

Which of the following would double the amount of current flowing through a piece of metal wire? Which of the following would do

uble the amount of current flowing through a piece of metal wire? Quarter the voltage across it. Quadruple the voltage across it. Double the voltage across it. Halve the voltage across it.
Physics
1 answer:
Strike441 [17]3 years ago
6 0

Answer:

Double the voltage across it.

Explanation:

From Ohm's Law, we have the relation, as follows:

Voltage = (Current)(Resistance)

If we keep the resistance of the wire as a constant, then we get the relation between voltage and current as follows:

Voltage = (Constant) * Current

Voltage α Current

This shows that the current has a direct proportionality with the voltage. So, if we want to double the current flowing through a piece of metal wire, then we must <u>Double the voltage across it.</u>

It can be confirmed by formula as:

V = IR

Now, doubling the voltage:

V' = 2 IR

substituting value of IR = V:

V' = 2 V

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Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

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The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

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m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

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\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

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