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2 years ago

8

The frequency of a certain sound is 440 Mz. What is the wavelength of this sound when the temperature of the air is (a) 20°C; (b

) 30°C

1 answer:

Serggg [28]2 years ago

3 0

Answer:

Explanation:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for

λ

:

λ

=

v

f

Let's plug in our given values and see what we get!

λ

=

340

m

s

440

s

−

1

λ

=

0.773

m

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Answer:

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Explanation:

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The force is attractive if the two charges have opposite sign, and repulsive if the two charges have same sign.

In this problem, we have:

$q_1 = -1 \mu C = -1 \cdot 10^{-6}C$ located at $x_1=-3 m$

$q_2=+9 \mu C = +9\cdot 10^{-6}C$ located at $x_2=0 m$

$q_3 = -5 \mu C = -5\cdot 10^{-6} C$ located at $x_3=+3 m$

The force between charge 1 and charge 2 is:

$F_{12}=k\frac{q_1 q_2}{(x_2-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(+9\cdot 10^{-6})}{3^2}=0.0090 N$

And since the two charges have opposite sign, the force is attractive, so the force on charge 1 is towards the right.

The force between charge 1 and charge 3 is:

$F_{13}=k\frac{q_1 q_3}{(x_3-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(5\cdot 10^{-6})}{6^2}=0.0012 N$

And since the two charges have same sign, the force is repulsive, so the force on charge 1 is towards the left.

Therefore, the net force on charge 1 is:

$F=F_{12}-F_{13}=0.0090-0.0012 = 0.0078 N$

towards the right.

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https://docs.googl e.com/presentation/d/e/2PACX-1vQ2d6GrYHS0WmSfvZlgMZ5dOH9UuHobdUDdWLernNTXKuH-DdWzkDsf6CBBqBbE1GGXNQ_2HmHncUHw

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