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2 years ago

Suggest two uses of pulleys.

Physics

1 answer:

Marizza181 [45]2 years ago

8 0

Answer: Elevators use pulleys to function.

A cargo lift system that allows for items to be hoisted to higher floors is a pulley system.

Explanation:

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Kobotan [32]

Answer:

Explanation:

Cause a monster truck don

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2 years ago

How do you find the kinetic energy of an object?

Sergeeva-Olga [200]

Answer:

KE = 1/2 * m * $v^{2}$

Explanation:

use the formula:

KE = 1/2 * m * $v^{2}$

KE = kinetic energy in joules (J)

m = mass in kg

v = velocity in m/s

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2 years ago

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In a certain region of space, the electric field is zero. from this fact, what can you conclude about the electric potential in

eduard

The answer is that it is constant. The relation between electric field and electric potential is given as, E= -gradient (V). The E, the partial rate of change of Electric potential, in the equation implies that the V, the partial differential of the potential of the three-dimensional space (assuming it is considered) is constant.

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2 years ago

Where does digestion begin?

Dafna11 [192]

Answer:

Digestion begins in the mouth, when you chew.

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2 years ago

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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

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

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3 years ago