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mafiozo [28]
3 years ago
7

A uniform electric field exists in the region between two oppositely charged plane-parallel plates. An electron is released from

rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 3.20 cm distant from the first, in a time interval of
a. Find the magnitude of this electric field.
b. Find the speed of the electron when it strikes the second plate
Physics
1 answer:
Zigmanuir [339]3 years ago
5 0

Answer:

Explanation:

  • given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s
  • q = 1.6 x 10_19C
  • using S = at^2/2
  • acceleration = 0.032 X 2 /(1.30×10−8)^2

a = 3.79 x 10^14m/s^2

  • From F = ma
  • F = qE
  • ma = qE

E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19

E = magnitude of this electric field. = 2156.3N/C

b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as

= 2 X 3.79 x 10^14 X 0.032

= 4.92 X 10^6m/s

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Alika [10]

(a) 1.01 cm

First of all, we need to find the spring constant of the spring.

The force initially applied to the spring is equal to the weight of the block hanging on it:

F=mg=(4.32)(9.8) = 42.3 N

where m = 4.32 kg is the mass of the block and g = 9.8 m/s^2 is the acceleration of gravity.

When this force is applied, the spring stretches by

x=2.92 cm = 0.0292 m

We can find the spring constant by using Hooke's law:

F=kx

where k is the spring constant. Solving for k,

k=\frac{F}{x}=\frac{42.3}{0.0292}=1448.6 N/m

Later, the first object is removed and another object of mass

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F'=m'g=(1.50)(9.8)=14.7 N

So, if we use Hooke's law again, we can find the new stretching of the spring:

x'=\frac{F'}{k}=\frac{14.7}{1448.6}=0.0101 m = 1.01 cm

(b) 1.16 J

The work that must be done on the spring is equal to the elastic potential energy that would be stored in the spring, therefore:

W=\frac{1}{2}kx^2

where we have

k = 1448.6 N/m is the spring constant

x = 4.00 cm = 0.04 m is the new stretching

Solving the equation, we find the work that must be done by the external force:

W=\frac{1}{2}(1448.6)(0.04)^2=1.16 J

6 0
3 years ago
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In-s [12.5K]

The correct option is (b) Sea breeze

During the day, susan notices that the wind is blowing onshore at the beach called sea breeze.

What is sea breeze?

  • Any wind that flows from a big body of water onto or onto a landmass is called a sea breeze or an onshore breeze.
  • Sea breezes form as a result of changes in air pressure brought on by the different heat capacities of water and dry land. Sea breezes are therefore more confined than prevailing winds.
  • A sea wind is frequently seen along coasts after sunrise because land warms up far more quickly than water does when exposed to solar radiation.
  • The sea wind front is significant because it can serve as a catalyst for afternoon thunderstorms and provide welcome cooling along the coast.

Learn more about the sea breeze with the help of the given link:

brainly.com/question/13015619

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kolbaska11 [484]

Explanation:

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Answer:

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