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3 years ago

Which runner has greater kinetic energy: a 45 kg runner moving at a speed of 7 m per second or a 93 kg runner moving at a

Physics

1 answer:

bagirrra123 [75]3 years ago

7 0

Kinetic Energy = (1/2) (mass) (speed)

First runner: KE = (1/2) (45kg) (49 m/s) = 1,102.5 Joules

Second runner: KE = (1/2) (93kg) (9 m/s) = 418.5 Joules

The <em>first runner </em><em>has 163</em>% more kinetic energy than the second runner has.

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Why are special techniques and extinguishing agents are required to fight combustible metals fires?

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3 years ago

A light ray transfers from more to less dense medium at certain condition , if we repeat this experiment by increasing the angle

evablogger [386]

Answer:

The correct option is;

Still constant

Explanation:

The relative refractive index ₁n₂ between the two medium can be as follows;

$_1n_2 = \dfrac{The \ speed \ of \ light \ in \ medium \ 1}{The \ speed \ of \ light \ in \ medium \ 2} = \dfrac{c_1}{c_2}$

Therefore, given that the speed of light in medium 1 is constant and the speed of light on medium 2 is also constant, the relative refractive index ₁n₂ = c₁/c₂ is always constant.

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3 years ago

what is the energy of an electromagnetic wave that has a frequency of 8.0 x 10^15 Hz? Use the equation...

Katarina [22]

(C)

Explanation:

$E = hf = (6.626×10^{-34}\:\text{J•s})(8.0×10^{15}\:\text{Hz})$

$= 5.3×10^{-18}\:\text{J}$

4 0

3 years ago

An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to

Amiraneli [1.4K]

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

$\theta_1$ is the angle of incidence (in the first medium)

$\theta_2$ is the angle of refraction (in the second medium)

Let's now consider a situation in which

$n_1 > n_2$

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

Where $\frac{n_1}{n_2}$ is a number greater than 1. This means that above a certain value of the angle of incidence $\theta_1$ , the term on the right can become greater than 1. So this would mean

$sin \theta_2 > 1$

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

$\theta_c =sin^{-1}(\frac{n_2}{n_1})$