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3 years ago

11

PS Final Exam

1 answer:

ivolga24 [154]3 years ago

4 0

A rolling friction jdisns

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What is the name of the line drawn perpendicular to the surface where a light ray strikes?

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It is the normal line

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3 years ago

If you decide you want to meet someone you met online, what should you do first?

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3 years ago

How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and

Dovator [93]

Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})$

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{9}{40}$

$f=4.44\ cm$

When , R₁= -4, R₂ = 8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{3}{40}$

$f=-13.33\ cm$

When , R₁= 4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{3}{40}$

$f=13.33\ cm$

When , R₁= -4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{9}{40}$

$f=-4.44\ cm$

Hence, The lenses with different focal length are four.

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3 years ago

Why was the microscope and its improvement an important part of the development of the cell theory?

Tresset [83]

Answer:it’s is part of the cell theory because they where studying cells and to see it you need a microscope

Explanation:basically in the answer area

7 0

3 years ago

Read 2 more answers

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

4 years ago