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Yakvenalex [24]
4 years ago
7

Unit of energy density of electric field is a) NC-3 b) JC-3 c) Jm-3 d)JF

Physics
2 answers:
erik [133]4 years ago
8 0

Answer:

the answer is c

Masteriza [31]4 years ago
5 0
The answer is C being JM-3
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How much time would it take to exert 700W of power while doing 700J of work?​
Valentin [98]

Answer:

The answer to your question is time = 1 s

Explanation:

Data

time = ?

Power = 700 W

Work = 700 J

Power is defined as the work done per unit of time.

Formula

Power = Work / Time

-Solve for time

Time = Work/Power

-Substitution

Time = 700 / 700

-Result

Time = 1 s

Conclusion

In 1 second and with 700 J there will be produced 700 watts

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To pull a nail out of a wood board a carpenter does 1000 J of work. The hammer he uses does 835 J of work. What is the efficienc
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S A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a frictionless, horizontal tr
Elina [12.6K]

The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

<h3>What is period of oscillation?</h3>

This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.

The period of oscillation is given as T

T = 2 * pi * sqrt ( m / k )

where

m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:

m1 = mass of the block

m2 = mass pf the spring

k = force constant of the spring

including the two masses to the period gives

T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

Read more on period of oscillation here: brainly.com/question/22499336

#SPJ4

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2 years ago
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