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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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The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-

                           V∝ R            [ where v is the rate of diffusion and r is the ratio of surface area to volume]

As per the question,the ratio of surface area to volume for a sphere is given 0.08m^{-1}

The surface area to volume ratio for right circular cylinder is given 2.1m^{-1}

Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.

Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.


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The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
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Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

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