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RUDIKE [14]
3 years ago
15

Different types of waves travel at different speeds. For example, sound waves travel more slowly than light waves. Have you ever

been in a big empty room, or in a cave, and heard an echo? Echos happen because sound waves take time to travel to the wall and bounce back to your ears. But light is different. Light is much faster than sound. Light is the fastest moving thing in the universe. Because light waves and sound waves travel at different speeds, you can figure out how far you are from a storm by watching for lightning and listening for thunder. Lightning and thunder are actually the same thing! Thunder is just the sound that lightning makes. Because light is so quick, you see the lightning flash first. The sound travels more slowly, so you hear the thunder after. Thunder and lightning represent two kinds of waves: sound and light.
How do lightning and thunder illustrate the relationship between light and sound waves?
A Light moves faster than sound, so you hear thunder after you see lightning.
B Sound moves faster than light, so you see lightning after you hear thunder.
C Lightning and thunder are the same, so light and sound waves move together at the same time.
D Sound moves light, so when you hear thunder you know lightning will move toward you.
Physics
1 answer:
trapecia [35]3 years ago
3 0

Answer:

A). Light moves faster than sound, so you hear thunder after you see lightning.

Explanation:

The lightning and thunder demonstrate that 'light travels faster than sound' which clearly portrays the relationship between the light and the soundwaves. When we see lightning in the sky before hearing the sound of thunder as the speed of light much higher than the speed of sound. Sound travels ~ 343 m/s or 1235 km/hr while light travels 300000 km/s. This difference in speed prevents us from considering lightning and thunder as the same thing. The 'sound is a pressure wave and therefore, takes time to travel and bounce back to our ears.' Thus, we happen to witness lightning first. Hence, <u>option A</u> is the correct answer.

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Answer:

a) t = \sqrt{\frac{h}{2g}}

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

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

Where:

y_{0_{1}}: is the initial height = h

v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

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

Where:

y_{0_{2}}: is the initial height of ball 2 = 0

y = v_{0_{2}}t - \frac{1}{2}gt^{2}    (2)

By equating equation (1) and (2) we have:

h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}

t=\frac{h}{v_{0_{2}}}

Where the initial velocity of the ball 2 is:

v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh

Since v_{f_{2}}^{2} = 0 at the maximum height (h):

v_{0_{2}} = \sqrt{2gh}

Hence, the time when they pass each other is:

t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}

b) When they are passing the speed of each one is:

For ball 1:

v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

The minus sign is because ball 1 is going down.

For ball 2:

v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

For ball 2:

y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!                  

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