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2 years ago

What is the true answer in this question getting a good night sleep

Physics

2 answers:

ohaa [14]2 years ago

6 0

Vas happenin!!!

It’s a yes,

I’d you get a good sleep you can do work and focus

Hope this helps

-Zayn Malik

Alexxx [7]2 years ago

4 0

Getting a good night sleep can benefit you in a lot of ways. Also that’s not a question.

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When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil

Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

$2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}$

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

$2t = 2\lambda_{film}$

$t = \lambda_{film}$

The index of refraction of soap is given, then

$\lambda_{film} = \frac{\lambda_{vacuum}}{n}$

Combining the results of all steps we get

$t = \frac{\lambda_{vacuum}}{n}$

Rearranging, we find

$\lambda_{vacuum} = tn$

$\lambda_{vacuum} = (278)(1.33)$

$\lambda_{vacuum} = 369.74nm$

4 0

2 years ago

We can use energy principles to make ____ predictions.

stira [4]

Compatible and speedy

5 0

3 years ago

A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball i

scoray [572]

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

The initial height of the cliff

yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

0 = 45.0 + 20×t –1/2×9.8×t²

0 = 45 +20t –4.9t²

Solving quadratically or by using a calculator,

t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

So this is the total time it takes for the ball to reach the ground from the height it was thrown.

The distance traveled by the woman is

s = vt

Given the speed of the woman v = 6.00m/s

Therefore

s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.

3 0

2 years ago

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$