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3 years ago

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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Explanation:

Given

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3 years ago

An observer is approaching at stationary source at 17.0 m/s. Assuming the speed of sound is 343 m/s, what is the frequency heard

UNO [17]

Answer:

the frequency heard by the observer is equal to 2677 Hz

Explanation:

given,

velocity of the observer = 17 m/s

speed of the sound = 343 m/s

velocity of the source = 0 m/s

frequency emitted from the source = 2550 Hz

$f = f_0(\dfrac{v-v_0}{v-v_s})$

$f = 2550\times (\dfrac{343+17}{343-0})$

velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz

hence, the frequency heard by the observer is equal to 2677 Hz

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3 years ago

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Then the total vertical distance is ...

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3 years ago

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Pepsi [2]

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<h3>Pythagoras theorem in brief :</h3>

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