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UNO [17]
2 years ago
15

A ball is launched from ground level at 20 m/s at an angle of 40° above the

Physics
1 answer:
DedPeter [7]2 years ago
8 0

(a) The ball's height <em>y</em> at time <em>t</em> is given by

<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :

0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )

<em>t</em> = 0   or   (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 <em>g t</em>

<em>t</em> = (40 m/s) sin(40º) / <em>g</em>

<em>t</em> ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So

0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>

where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :

<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)

<em>y</em> ≈ 8.4 m

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g n diffraction, the formula for minima is given by a times s i n (theta )equals m lambda, where a is the width of the slit, the
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Answer:

θ = 22.2

Explanation:

This is a diffraction exercise

        a sin θ = m λ

The extension of the third zero is requested (m = 3)

They indicate the wavelength  λ = 630 nm = 630 10⁻⁹ m and the width of the slit  a = 5 10⁻⁶ m

         sin  θ = m λ / a

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          θ = sin⁻¹  0.378

         

to better see the result let's find the angle in radians

          θ = 0.3876 rad

let's reduce to degrees

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During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshi
vladimir2022 [97]

Answer:

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Explanation:

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wgere

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t = time = 36 ms

as can be seen, the above equation calls up the given variable as a function of the required variable thus

S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m

At 60g, a person will travel a distance of 0.38 m before coing to a complete stop

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