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sattari [20]
3 years ago
5

If the electric field between the plates has a magnitude of 1.4×105 V/m , what is the potential difference between the plates?

Physics
1 answer:
Bogdan [553]3 years ago
5 0

Answer: V = Ed

V = 107.8 Volts

The distance between plate is missing, below is a possible complete question:

A Parallel-plate Capacitor Has Plates Separated By 0.77 Mm,If the electric field between the plates has a magnitude of 1.4×105 V/m , what is the potential difference between the plates?

Explanation:

Potential difference V = electric field E × distance d

V = Ed

E = 1.4 × 10^5 V/m

d = 0.77mm = 0.77 × 10^-3 m

V = 1.4 × 0.77 × 10^(5-3) Volt

V = 1.078 × 10^2 Volts

V = 107.8 Volts

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<u>Explanation:</u>

Frequency is the measure of repetition of same thing a certain number of times. So frequency is inversely proportional to the wavelength. As wavelength is distance between two successive crests or troughs in a sound wave.

And frequency is the completion of number of cycles in a given time in sound waves. The frequency and wavelength are inversely proportional to each other with velocity of sound being the proportionality constant.

Thus, here the speed of sound is given as 343 m/s, the wavelength of the note is also given as 2.62 m, then frequency will be as follows:

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3 years ago
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xenn [34]

Correct answer choice is :


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Explanation:


Fluids have a particular temperature at which they convert into solids, identified as their freezing point. In theory, the melting point of a solid should be the same as the freezing point of the liquid. In practice, small variations among these measures can be seen. The freezing point of a matter is the same as that substance's melting point. At this distinct temperature, the substance can exist as either a solid or a liquid. At temperatures below the freezing/ melting point, the substance is a solid.


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