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ladessa [460]
3 years ago
9

The man who discovered that even individual light particles have wave characteristics was:

Physics
2 answers:
wolverine [178]3 years ago
4 0
I am almost positive that it was, ISSAC NEWTON
Len [333]3 years ago
3 0

Answer:

c. Taylor

Explanation:

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Two truckers are traveling directly away from each other at the same speed. If one trucker sounds her horn at a frequency of 221
8_murik_8 [283]

Answer:

v = 8.8 m /s

Explanation:

For listener and source going away from each other the formula of Doppler effect is as follows

\frac{f}{f_0} = \frac{V-v}{V+v}

V is velocity of sound , v is velocity of listner and source of sound

f₀ is apparent frequency and f is real frequency

V = 343 , v = ? ,f = 210 , f₀ = 221

Put these value in the relation above

[tex]\frac{210}{221} = \frac{343-v}{343+v}[/tex]

v = 8.8 m /s

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Martin Cooper invented the cell phone
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A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. What is the maximum height h that it achieves, and w
sveta [45]

Answer:

A)     t = 0.40816 s , y = 0.916 m

Explanation:

A) For this problem we use the kinematic relations

           v = v₀ - g t

the highest point zero velocities (v = 0)

           t = (v₀-v) / g

           t = (4 - 0) / 9.8

           t = 0.40816 s

to calculate the height let's use

          v² = v₀² - 2 g y

          y = vo2 / 2g

           y = 4 2 / (2 9.8)

          y = 0.916 m

To find the number of photos, we can use a direct proportions rule, if you take 30 photos in a second in 0.40816 s how many photos does it take

          # _photos1 = 0.40916 (30/1)

          # _photos1 = 12

yes i take 120 fps

          #_fotod = 0.40916 (120/1)

          #photos = 5.87 10³

 

B) The ball is released from a latura h how long it takes to reach the floor

           v² = v₀² + 2 g y

where the initial velocity is zero and the velocity with which the expert leaves is equal to the velocity with which v = 4 m / s leaves

            v² = 2gy

             v = √ (2 9.8 0.916)

             v = √ (2.1397 101)

             v = 4.6257 m / s

c) we ask us for the time for latura

             y = L / 2

             y = 0.916 / 2

             y = 0.458 m

now we can use the formula

             y = v₀ t - ½ g t²

           0.458 = 4.00 t - ½ 9.8 t²

            4.9 t² - 4t + 0.458 = 0

            t² -0.8163 t +0.09346 = 0

we solve second degree execution

           t = [0.8163 ±√ (0.8163² - 4 0.09346)] / 2

           t = [0.8163 ± 0.540] / 2

           t₁ = 0.678 m

           t₂ = 0.2763 m

the shortest time is for when the ball goes up and the longest when it goes down

D) the graph of vs Vs is expected to be a closed line

and the graph of position versus time a parabola

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