All categories

3 years ago

What is the wavelength in nm of a light whose first order bright band forms a

Physics

1 answer:

scoundrel [369]3 years ago

3 0

The wavelength of the first order bright band light light is 714 nm .

Explanation:

We have to find the wavelength of the first order brightness of a light. Here we are using Huygen's principle of light.

The formula is

nλ =d sinθ

where, n is the order of maximum

λ is the wavelength of light

d is the distance between the lines on diffraction grating.

θ is the angle.

For the given equation n is 1 because the problem states that the light forms 1st order bright band

λ is unknown.

d = $\frac{1}{700,000}$ or 0.0000014 m

sin (30) = 0.5

so,

1(λ) = (0.0000014)(0.5)

= 0.0000000714

= 714 nm

Thus, The wavelength of the first order bright band light light is 714 nm .

You might be interested in

What planet has a visible mark referred to as the the Great Red Spot?

vivado [14]

Jupiter that is the answer

good luck

6 0

2 years ago

Can you help me with this??

s2008m [1.1K]

Answer:

i want to say flip the coins but im not really sure sry

Explanation:

3 0

3 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago

A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet

puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

T2 = 453 - 273 = 180° C

3 0

3 years ago

You are on a school bus driving north at 20 m/s. You walk toward the back of the bus with a velocity of 6 m/s. What is your velo

Ganezh [65]

I answered the question but it got deleted?? why?

3 0

3 years ago