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Lesechka [4]
3 years ago
10

A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction

is mu_k. During time interval t, the sled moves a distance s, starting from rest.Find the average velocity vavg of the sled during that time interval.Express your answer in terms of the given quantities and, if necessary, appropriate constants. You may or may not use all of the given quantities.
Physics
1 answer:
lorasvet [3.4K]3 years ago
4 0

Answer:

The average velocity of the sled is vavg = s/t.

Explanation:

Hi there!

The average velocity is calculated as the traveled distance over time:

vavg = Δx/Δt

Where:

vavg = average velocity.

Δx = traveled distance.

Δt = elapsed time.

We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:

vavg = s/t

Have a nice day!

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Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul
Yuki888 [10]

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

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

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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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