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garri49 [273]
3 years ago
11

Which term describes an area of low density in a longitudinal wave?

Physics
2 answers:
LiRa [457]3 years ago
6 0

Answer:

Explanation:

Low density is an area of rarefaction.

kicyunya [14]3 years ago
6 0

Answer: rarefaction

Explanation: correct answer on a p e x

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Thermal energy is added to four identical samples of water what increases in each sample
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Thermal energy is added to four identical<span> 1.0 kg </span>samples of water<span> at room temperature. Which of the following </span>increases in each sample<span>? average charge of an electron; average density of </span>a<span> nucleus; average mass of </span>a<span> proton; average speed of </span>a<span> molecule. Your answer: -. Answer: D - average speed of </span>a<span>molecule.</span>
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2 years ago
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Which of the following is the most important factor in determining climate
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Answer:

temperature

The two most important factors in the climate of an area are temperature and precipitation. The yearly average temperature of the area is obviously important, but the yearly range in temperature is also important.

Explanation:

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3 years ago
Jack and Jill have made up since the previous HW assignment, and are now playing on a 10 meter seesaw. Jill is sitting on one en
Airida [17]

Answer: 3 m.

Explanation:

Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by  gravity acting on both children  must be 0.

As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.

If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):

mJill* 5m -mJack* d = 0

60 kg*5 m -100 kg* d =0

Solving for d:

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6 0
3 years ago
The US Environmental Protection Agency issues a daily report for pollution levels called the __________.
VARVARA [1.3K]
I think it’s D lol i’m not sure
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3 years ago
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Write an expression for a harmonic wave with an amplitude of 0.19 m, a wavelength of 2.6 m, and a period of 1.2 s. The wave is t
zlopas [31]

Answer:

y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})

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As we know that the wave equation is given as

y = A sin(\omega t - k x + \phi_0)

now we have

A = 0.19 m

\lambda = 2.6 m

so we have

k = \frac{2\pi}{\lambda}

k = \frac{2\pi}{2.6}

k = 2.42  per m

also we have

T = 1.2 s

so we have

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{1.2}

\omega = 5.23 rad/s

now we know that at t = 0 and x = 0 wave is at y = 0.19 m

so we have

\phi_0 = \frac{\pi}{2}

so we have

y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})

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