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Goryan [66]
3 years ago
9

A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person boxing through

the window for 200 ms as it passes by the 2.00 m height of the window. From what height above the top of the window was the ball dropped?
Physics
1 answer:
Irina18 [472]3 years ago
5 0

Answer:

4.14 m

Explanation:

In the last leg of the journey the ball covers 2 m in 2ms or 0.2 s .

Let in this last leg , u be the initial velocity.

s = ut + 1/2 g t²

2 = .2 u + .5 x 9.8 x .04

u = 9.02 m /s .

Let v be the final velocity in this leg

v² = u² + 2 g s

v² = (9.02)² + 2 x 9.8 x 2

= 81.36 +39.2

v = 10.97 m / s

Now consider the whole height from where the ball dropped . Let it be h.

Initial velocity u = 0

v² = u² +2gh

(10.97 )² = 2 x 9.8 h

h = 6.14 m

Height from window

= 6.14 - 2m

= 4.14 m

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Answer:

241.7 s

Explanation:

We are given that

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Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

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Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

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3.626 m/s

Explanation:

v=d/t

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2. 0.86/0.2 = 4.3 m/s

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5. 3.32/0.8 = 4.15 m/s

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7. 4.79/1.2 = 3.99 m/s

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9. 6.15/1.6 = 3.84 m/s

10. 6.76/1.8 = 3.76 m/s

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12. 7.92/2.2 = 3.6 m/s

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3 years ago
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Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

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<u>Where:</u>

θ: is the angle of resolution of a circular aperture

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d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

\lambda = \frac{c}{f}

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c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m            

Hence, the distance is:

d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

 

I hope it helps you!

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