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goldfiish [28.3K]
3 years ago
6

A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just

above the surface of the sheet?
Physics
1 answer:
Alex73 [517]3 years ago
7 0

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

E = \frac {\sigma }{2\epsilon _{0}}

E = \frac{9\times 10^{-6}}{2\times 8.854\times 10^{-12}}

E = 5.08 x 10^5 N/C

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PilotLPTM [1.2K]

Answer:

freezing point and melting point

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3 years ago
A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea
viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}                           (1)

time taken = 5 years

We know that :

time = \frac{distance}{speed}

5 = \frac{L''}{v}                                                      (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}

Solving the above eqn:

t'' = 20.56 years

4 0
3 years ago
Zoe is shown two mystery boxes that are both 8,000cm3. Her teacher tells her that one mystery box is filled with rocks and the o
krek1111 [17]

Answer:

The box of rocks will have depression which can be seen without touching the box.

Explanation:

The density of rocks is very large as compared with napkins. So, the weight of the rocks will be much more greater than that of napkins.

As both boxes have same volume the heavier box will show depression on the lower surface as compared to the lighter box. So, the box of rocks will have depression which can be seen without touching the box.

5 0
3 years ago
I just need the answer to number 1 (with work please) <br> (Newton’s Second Law)
Valentin [98]

Answer:

wish I could help

Explanation:

I been rereading this and I can't solve it lemme go ask people in ma house real quick

8 0
3 years ago
A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch
Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

8 0
3 years ago
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