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Len [333]
3 years ago
15

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see

the drawing). The plates are separated by a distance of 1.1 cm, and the electric field within the capacitor has a magnitude of 2.2 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?
Physics
1 answer:
siniylev [52]3 years ago
6 0

Answer:

3.88 * 10^(-15) J

Explanation:

We know that the Potential energy of the electron at the beginning of its motion is equal to the Kinetic energy at the end of its motion, when it reaches the plates.

First, we get the potential and potential energy:

Electric potential = E * r

E = electric field

r = distance between plates

Potential = 2.2 * 10^6 * 0.011

= 2.42 * 10^4 V

The relationship between electric potential and potential energy is:

P. E. = q*V

q = charge of electron = 1.602 * 10^(-19) C

P. E. = 2.42 * 10^4 * 1.602 * 10^(-19)

P. E. = 3.88 * 10^(-15) J

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Veronika [31]

The object's final velocity, given the data is 10.5 rad/s

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

  • a is the acceleration
  • v is the final velocity
  • u is the initial velocity
  • t is the time

<h3>How to determine the final velocity</h3>

The following data were obtained from the question

  • Initial velocity (u) = 1.5 rad/s
  • Acceleration (a) = 0.75 rad/s²
  • Time (t) = 12 s
  • Final velocity (v) = ?

The final velocity can be obtained as follow:

a = (v – u) / t

0.75 = (v – 1.5) / 12

Cross multiply

v – 1.5 = 0.75 × 12

v – 1.5 = 9

Collect like terms

v = 9 + 1.5

v = 10.5 rad/s

Thus, the final velocity of the object is 10.5 rad/s

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6 0
2 years ago
A 850-kg sports car accelerates from rest to 95 km/h in 6.8 s .
swat32

From the calculations, the power expended is 43650 W.

<h3>What is the power expended?</h3>

Now we can find the acceleration from;

v = u + at

u = 0 m/s

v =  95 km/h or 26.4 m/s

t =  6.8 s

a = ?

Now

v = at

a = v/t

a = 26.4 m/s/ 6.8 s

a = 3.88 m/s^2

Force = ma = 850-kg * 3.88 m/s^2 = 3298 N

The distance covered is obtained from;

v^2 = u^2 + 2as

v^2 = 2as

s = v^2/2a

s = (26.4)^2/2 * 3.88

s = 696.96/7.76

s = 90 m

Now;

Work = Fs

Work =  3298 N * 90 m = 296820 J

Power =  296820 J/ 6.8 s

= 43650 W

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2 years ago
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borishaifa [10]

Answer:

average velocity include total displacement whereas average speed include total distance

6 0
3 years ago
What is The distance traveled by a vehicle in 12 minutes,if it’s speed is 35km/h
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Answer: 7 km

Explanation: 12 minutes is 1/5 of an hour. 35/5=7

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A new pest is found that only feeds on corn plant leaves. Which promoter would you choose if the modified gene produces toxins t
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