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Sav [38]
3 years ago
7

What is the speed of a bobsled whose distance-time graph indicates that it traveled 122m in 27s?

Physics
1 answer:
Stolb23 [73]3 years ago
7 0
Speed = distance/time
speed= 122÷27=4.52m/s (3sf)
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Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second
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To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

I' = \frac{I_0}{2} cos^2 \theta

Where

\theta =Angle From vertical of the axis of the polarizing filter

I_0 = Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

I = \frac{I_0}{2}

Replacing we have that

I = \frac{370}{2}

I = 185W/m^2

Re-arrange the equation,

I'= \frac{I_0}{2}cos^2\theta

Re-arrange to find \theta

cos^2\theta = \frac{2I'}{I_0}

cos^2\theta = \frac{2*138}{370}

\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})

\theta = 0.5282rad

\theta = 30.27\°

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

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Motor 1 and motor 2 each lift objects of equal weight. They both have the same power rating but motor 2 lifts the weight faster.
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Answer:

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Answer:

44.3 m/s

Explanation:

Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.

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The parameters given are:

Height H = 100m

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Let g = 9.8m/s^2

Using third equation of motion

V^2 = U^2 + 2gH

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V^2 = 2 × 9.8 × 100

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V = 44.27 m/s

Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately

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