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3 years ago

5

A student measures the speed of sound by echo destiny classes hands and then measures the time to hear the echo his distance to

the wall is 300 m The time delay between clap an echo is 1.5 seconds. Calculate the speed of sound

1 answer:

777dan777 [17]3 years ago

6 0

Explanation:

∆x=300 m×2

∆t=1.5 s

v=∆x/∆t → v=2×300/1.5 = 400 m/s

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Pluto has been reassigned and is now a dwarf planet. Why did scientists think this reassignment was necessary? If you were a sci

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A student observes that it is hard to hear music underwater in a pool. They state that the sound is always muffled. They

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Answer:

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Explanation:

The answer is false.

The speed of the sound in water is faster when compared to the speed of sound in air. This is because, the particles in air is loosely packed and are far from each other as compared to water or liquid.

The water particles are close to each other than air particles, so water particles are able to transmit the vibrations of the sound faster than the air particles.

Therefore sound waves travels faster in water than in air.

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3 years ago

An object of mass 6.00 kg falls with an aceleration of 8.00 m/s2. The magnitude of air resitance must be ____ N

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Using Newton's Second Law, we can find the air resistance. We know the net force is equal to mass times acceleration. $F_{net} = m*a = (6.00kg)(8.00 \frac{m}{s^2}) = 48N 

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48N = (6.00kg)(9.81m/s^2) - F_{d} 

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3 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

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4 years ago