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2 years ago

Could anyone help me for this question please!! Really emergency!!!

Physics

1 answer:

kodGreya [7K]2 years ago

5 0

16/9 m/s^2
negative 4/3 m/s^2
14 m/s
the last one is too detailed to do in my head while on the bus; sorry

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What happens to light when it travels from air into water

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Answer:

Water is more dense than air. When water goes through a denser thing, the light is "bent" more towards the "normal" which is a straight, vertical line.

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2 years ago

What component of an atom determines what element it is?

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The number of protons in the element.

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3 years ago

Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul

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Answer:

$63.4^{\circ}$

Explanation:

When unpolarized light passes through the first polarizer, the intensity of the light is reduced by a factor 1/2, so

$I_1 = \frac{1}{2}I_0$ (1)

where I_0 is the intensity of the initial unpolarized light, while I_1 is the intensity of the polarized light coming out from the first filter. Light that comes out from the first polarizer is also polarized, in the same direction as the axis of the first polarizer.

When the (now polarized) light hits the second polarizer, whose axis of polarization is rotated by an angle $\theta$ with respect to the first one, the intensity of the light coming out is

$I_2 = I_1 cos^2 \theta$ (2)

If we combine (1) and (2) together,

$I_2 = \frac{1}{2}I_0 cos^2 \theta$ (3)

We want the final intensity to be 1/10 the initial intensity, so

$I_2 = \frac{1}{10}I_0$

So we can rewrite (3) as

$\frac{1}{10}I_0 = \frac{1}{2}I_0 cos^2 \theta$

From which we find

$cos^2 \theta = \frac{1}{5}$

$cos \theta = \frac{1}{\sqrt{5}}$

$\theta=cos^{-1}(\frac{1}{\sqrt{5}})=63.4^{\circ}$

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2 years ago

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Answer:

Explanation:

the electrons beams deflection radius will be halved.

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2 years ago

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To solve this problem we will apply the concepts related to the kinematic equations of linear motion. For this purpose we will define the speed as the distance traveled in a given period of time. Here the distance is equivalent to the orbit traveled around the earth, that is, a circle. Approaching the height of the aircraft with the radius of the earth, we will have the following data,

$R= 6370*10^3 m$