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Ierofanga [76]
3 years ago
9

How do Air molecules enable sound to travel from a radios speaker to your ears?

Physics
2 answers:
MatroZZZ [7]3 years ago
6 0
The radio frequencies push one air molecule that then bumps into a different air molecule.....which then hits another and another causing a line of crashing molecules that lead inside your ear and hits your ear drum causing it to vibrate which causes the sounds.
madreJ [45]3 years ago
6 0
The molecules bounce and colide and hit like dominoes and vibrate and we haer it
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The period of a carrier wave is T=0.01 seconds. Determine the frequency and wavelength of the carrier wave. a. f=10 Hz, λ=3E8 me
makvit [3.9K]

Explanation:

It is given that,

The period of the carrier wave, T = 0.01 s

Let f and \lambda are frequency and the wavelength of the wave respectively. The relationship between the time period and the frequency is given by :

f=\dfrac{1}{T}

f=\dfrac{1}{0.01}

f = 100 Hz

The wavelength of a wave is given by :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8}{100}

\lambda=3\times 10^6\ m

So, the  frequency and wavelength of the carrier wave are 100 Hz and 3\times 10^6\ m respectively. Hence, the correct option is (c).

6 0
3 years ago
A major difference between AC and DC electricity is that A) the current in AC electricity varies in magnitude and direction. B)
galina1969 [7]

Answer:

A) the current in AC electricity varies in magnitude and direction.

C) the voltage in AC electricity varies in magnitude and direction.

Explanation:

In DC current and voltage the direction of current will not change with time and it always remains the same.

So here in DC voltage and DC current the magnitude may change with time but the direction will always remain same

While in AC voltage and AC current the direction of AC will change with time

periodically.

So here magnitude and direction both will change in AC current and AC voltage.

so the correct answer is

A)     the current in AC electricity varies in magnitude and direction.

C) the voltage in AC electricity varies in magnitude and direction.

7 0
3 years ago
Read 2 more answers
The following is current scientific evidence supporting the nebular theory on the formation of the solar system.
Feliz [49]
A.the composition of the inner and outer planets, current observations of star formation, and the motion of the solar system I hope this helps
4 0
3 years ago
Read 2 more answers
What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?
zubka84 [21]

Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

Vp = 0.94 m/s

6 0
3 years ago
Read 2 more answers
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
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