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Tamiku [17]
3 years ago
14

How can we realize that light travel in straight line ?​

Physics
1 answer:
Norma-Jean [14]3 years ago
3 0

Answer:

It can be seen from the operation of pin-hole camera, formation of shadows and eclipse.

Explanation:

The phenomenon of light traveling in a straight line is known as rectilinear propagation of light.

One this evidence can be seen from the operation of pin-hole camera, which depends on rectilinear propagation of light

Also two natural effects that result from the rectilinear propagation of light are the formation of Shadows and Eclipse.  

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A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge
Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is \lambda and the net charge per unit length is 2 \lambda.

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let \lambda_i\ and\ \lambda_o are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

\phi=\dfrac{q_{enclosed}}{\epsilon_o}

E.A=\dfrac{q_{enclosed}}{\epsilon_o}

0=\dfrac{\lambda_i+\lambda}{\epsilon_o}

\lambda_i=-\lambda  

For outer surface,

\lambda_i+\lambda_o=2\lambda

-\lambda+\lambda_o=2\lambda

\lambda_o=3\lambda

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}

E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}

Hence, this is the required solution.

6 0
3 years ago
I didnt want my question public i made a mistake i want it taken down
mrs_skeptik [129]
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5 0
2 years ago
Electronegativity increases when atoms ___
notka56 [123]

your answer is.....

D. have a large atomic radius

although they also increase going from left to right so if D is incorrect, B might be your answer. it depends on context of the lesson.

8 0
3 years ago
What is the name of the hair-like structures on sponge cells that move back and forth to help move water, nutrients, and waste t
Dima020 [189]
I think it's B. Flagella

5 0
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An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

6 0
3 years ago
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