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Svetlanka [38]
3 years ago
6

The rms current in an ac current is 3.6 a. find the maximum current

Physics
1 answer:
LUCKY_DIMON [66]3 years ago
7 0
A peak = A Rms x Sq root 2

Therefore 3.6 x sq root of 2
A peak = 5.09
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Which of the following is same dimension quantity?
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Answer:

Find the dimension of each and every quantity in all the options to check whether they are the same or not. We can use any one formula of each identity to find its dimension.

Complete step by step solution:

To find the dimension of a quantity, we can use any formula related to that quantity but we will use the easiest ones to save time.

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F=maF=ma

Dimension of force =[M][LT−2]=[MLT−2]=[M][LT−2]=[MLT−2]

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p=mvp=mv

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Explain how energy allows a paper clip to be attracted to a magnet
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*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude
kvv77 [185]

Answer:

Approximately 3.86\; {\rm N} (given that the magnitude of this charge is -7.35\; {\rm \mu C}.)

Explanation:

If a charge of magnitude q is placed in an electric field of magnitude E, the magnitude of the electrostatic force on that charge would be F = E\, q.

The magnitude of this charge is q = 7.35\; {\rm \mu C}. Apply the unit conversion 1\; {\rm \mu C} = 10^{-6}\; {\rm C}:

\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}.

An electric field of magnitude E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}} would exert on this charge a force with a magnitude of:

\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

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