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scZoUnD [109]
3 years ago
15

A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next

to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed
Physics
1 answer:
zalisa [80]3 years ago
5 0

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

V=1.4*10^5m/s

Explanation:

From the question we are told that:

Electric field B=1.5*10N/C

Distance d=2 x 10^{-3}

At negative plate

Generally the equation for Velocity is mathematically given by

V^2=2as

Therefore

V^2=\frac{2*e_0E*d}{m}

V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}

V=\sqrt{19.2*10^9}

V=1.4*10^5m/s

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The engineer of a passenger train traveling at 25.0m/s sights a freight train whose caboose is 200m ahead on the same track. The
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No, there won't be a collision.

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James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s
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Answer:

Both of them reach the lake at the same time.

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    Substituting,

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Both of them reach the lake at the same time.

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3 years ago
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