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adoni [48]
3 years ago
5

Light has wavelength 600 nm in a vacuum. it passes into glass, which has an index of refraction of 1.5. what is the frequency of

the light inside the glass
Physics
1 answer:
Lelu [443]3 years ago
7 0

v = f lambda

in vac ... 3X10^8 = 600x10^-9xf

in glass speed slower, poss 2/3 that of vacuum

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Hi! The answer is ‘B’! Because the nucleus is found at the center and contains protons (positive charge) and neutrons (no charge)
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3 years ago
If 270 watts of power is used in 42 seconds, how much work was done<br>​
navik [9.2K]

Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

<u>Step 1:</u><em><u> Find the Formula</u></em>

P = W / T

<em><u> </u></em>

<u>Step 2: </u><u><em>Make W the subject of the equation.</em></u>

W = PT

<u>Step 3:</u><u> </u><u><em>Given.</em></u>

P = 270 Watts, T = 42 seconds

<u>Step 4:</u><u><em> Substitute these values into equation 2 .</em></u>

W = 270(42)

<u>Step 5:</u><u> </u><u><em>Simplify.</em></u>

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

4 0
3 years ago
The sun rotates once in 25.05 days. That is 601.2 hours. so the sun's rotational speed is 0.0017 rotations per hour. what is the
Margaret [11]

Answer:

v=2019.09\ m.s^{-1}

Explanation:

Given:

  • time taken by the sun to complete one revolution, t=601.2\ hr
  • radial distance of the sunspot, r=6.955\times 10^8\ m.s^{-1}

<u>Therefore, angular speed of rotation of sun:</u>

\omega=\frac{2\pi}{601.2\times 3600} \ rad.s^{-1}

<u>Now the tangential velocity of the sunspot can be given by:</u>

v=r.\omega

v=6.955\times 10^8\times \frac{2\pi}{601.2\times 3600}

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4 0
3 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

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3 years ago
What is one fun fact about both rubies and sapphires?
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Answer:

A ruby is measures a 9 on the Mohs Scale of Hardness, second only to diamond and matched with sapphire. This makes a ruby an extremely hard and durable gemstone.

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8 0
3 years ago
Read 2 more answers
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