I need help on 6, 7, 8, and 9

A resting object exerts pressure equal to (i point!

DIA [1.3K]

I think it’s the last one it’s weight divided by its contact area

4 0

3 years ago

White light (400–700 nm) is incident on a 600 line/mm diffraction grating. What is the width of the first-order rainbow on a scr

Tema [17]

Answer:

Δy=0.431m

Explanation:

Diffraction grating with split space d,to find the fringe position ym,we must to find the angle from

dSinα=mλ

A grating with N slits or lines per mm has slit spacing of

d=1/N

d=(1/600mm)

d=1.67×10⁻³mm

For 400nm wavelength:

α=Sin⁻¹(mλ/d)

$\alpha =Sin^{-1}(\frac{400*10^{-9} }{1.67*10^{-6}} )\\ \alpha =13.910^{o}$

And the position of first order lowest wavelength fringe on the screen is:

$y_{1}=Ltan\alpha_{1}\\y_{1}=2tan(13.910)\\ y_{1}=0.49445m$

For 700nm wavelength:

α=Sin⁻¹(mλ/d)

$\alpha =Sin^{-1}(\frac{700*10^{-9} }{1.67*10^{-6}} )\\ \alpha_{2} =24.83^{o}$

And the position of first order highest wavelength fringe on the screen is:

$y_{2}=Ltan\alpha_{2}\\y_{2}=2tan(24.83)\\ y_{2}=0.925595m$

The difference between the first order lowest and highest wavelength fringe is

Δy=(0.925595 - 0.49445)m

Δy=0.431m

6 0

3 years ago

According to the work-energy theorem, if work is done on an object, its potential and/or kinetic energy changes. Consider a car

Leni [432]

Answer:

The force of the car engine.

Explanation:

The work- energy theorem states that the work done on an object is equal to the change in its kinetic energy. Its expression is given by :

$W=\dfrac{1}{2}m(v^2-u^2)$

Also, W = F.d

$Fd=\dfrac{1}{2}m(v^2-u^2)$

Where

F is the force applied by the engine of car

d is the displacement

m is the mass of an object

u is the initial speed

v is the final speed

So, the force of the car engine increased the car’s kinetic energy. Hence, this is the required solution.

6 0

3 years ago

What should be the angle between the transmission axes of the polarizers if it is desired that one-tenth of the incident intensi

NNADVOKAT [17]

The angle between the transmission axes of the polarizers if it is desired that one-tenth of the incident intensity be transmitted, ∝ = 63.435°

We'll assume that $I_{0}$ represents the incident light's intensity. Two polarizers are provided to us, but the first polarizer's angle is not disclosed. The incident light in situations like this needs to be unpolarized. This is due to the fact that, regardless of angle, the transmitted intensity via the polarizer is reduced by half for unpolarized light:

$I_{1} =\frac{1}{2} I_{0}$

Then,

$I_{2} =\frac{1}{10} I_{0}$

By using Malu's law,

$I_{2} =I_{1} cos^{2} \alpha$

That is,

$cos^{2} \alpha = \frac{I_{2} }{I_{1} }$

In terms of $I_{0}$ , we get

$cos^{2} \alpha =\frac{0.1I_{0} }{0.5I_{0} }$

$cos^{2} \alpha = 0.2$

Now the angle is,

$cos\alpha =\sqrt{0.2}$

cos ∝ = 0.44721

$\alpha =cos^{-1} (0.44721)$

Then, ∝ = 63.435°

This angle is measured with respect to the first polarizer angle.

Learn more about the polarizers here: brainly.com/question/13008007

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7 0

2 years ago

A mass moves back and forth in simple harmonic motion with amplitude A and period T. (A) In terms of amplitude , through what di

elixir [45]

The distance the mass movement in the time T is 4A.

The amplitude is defined as the maximum displacement of the system from the equilibrium position.

The period refers to the time taken by the system to make one complete oscillation.The time it takes to go from a displacement of x=+A to the next x=+A.We have that the total distance covered by the system in one period T is 4 times the amplitude as 4A as It moves from x=+A to x=0, the distance covered is A.

When it moves from x=0 to x=-A, the distance covered is A+A=2AIf it, moves from x=-A to x=0, distance covered is 2A+A=3AWhen it moves from x=0 to x=+a, distance covered is 3A+A=4AHence, the distance does the mass move in the time T is 4A.

To learn more about distance, click here brainly.com/question/12530283

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4 0

1 year ago