Do compressional waves need a medium? If so which one would be the fastest ans slowest medium.

1 answer:

5 0

Yes, compression waves need a medium. without a medium there is nothing to compress, hence, no wave. A fast medium like a gas (air) is easy to compress and allows waves to move through it easily. a slow medium, like a liquid, is still pretty fast, but not as fast as air.

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An incident ray of light strikes water at an angle of 30°. The index of refraction of air is 1.0003, and the index of refraction

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Answer: It’s A

Explanation:

22 degrees

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2 years ago

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Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

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The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
$F=qvB \sin \theta$
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1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is $q=2.7 \mu C=2.7 \cdot 10^{-6}C$ . In this case, v and B are perpendicular, so $\theta=90^{\circ}$ , therefore we have:
$B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T$

2) In this second case, the angle between v and B is $\theta=55^{\circ}$ . The charge is now $q=42.0 \mu C=42.0 \cdot 10^{-6}C$ , and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
$F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N$

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A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 k. The temperature is decrease

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The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:

$P_2 = T_2 \frac{P_1}{T_1}=(101.8 K) \frac{5 atm}{254.5 K}=2 atm$