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kenny6666 [7]
4 years ago
8

Can you differentiate between the wavelength of visible light and the wavelength of a moving soccer ball

Physics
2 answers:
KIM [24]4 years ago
7 0

Answer

The wavelength of  a moving soccer ball is invisible using the naked eye.

Explanation

The wavelength of a moving soccer ball is not detectable with the naked eye because it is smaller than the wavelength of visible light. The human eye is only able to see about 17 frames per second. The wavelength of a moving soccer ball can be compared to the infinite value of the de Broglie wavelength where the Plank’s constant is so small.

natulia [17]4 years ago
5 0
A dramatic difference. Moving soccer ball does not have any wavelength. It's not a wave process at all. 
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A cup sits on a table. Due to its position, the potential energy of the cup is 3.00 joules. Ignoring frictional effects, if the
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Which of the following objects would absorb the most light?
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A an opanque object is compared to a black T shirt

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3 years ago
Two movers are pushing a large crate with a force of 60.0 n each. one pushes north, the other east. what is the equilibrant forc
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8 0
3 years ago
a stone is thrown horizonttaly from a cliff of a hill with an initial velocity of 30m/s it hits the ground at a horizontal dista
ELEN [110]

Answer:

a) Time = 2.67 s

b) Height = 35.0 m

Explanation:

a) The time of flight can be found using the following equation:

x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}   (1)

Where:

x_{f}: is the final position in the horizontal direction = 80 m

x_{0}: is the initial position in the horizontal direction = 0

v_{0_{x}}: is the initial velocity in the horizontal direction = 30 m/s

a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)

t: is the time =?  

By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:  

t = \frac{x_{f}}{v_{0}} = \frac{80 m}{30 m/s} = 2.67 s

b) The height of the hill is given by:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final position in the vertical direction = 0

y_{0}: is the initial position in the vertical direction =?

v_{0_{y}}: is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)            

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the hill is:

y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m  

I hope it helps you!

5 0
3 years ago
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