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faltersainse [42]
3 years ago
8

The diagram illustrates the movement of sound waves between an observer and a race car. As the race car drives away from the obs

erver, notes that the sound from the car gets quieter and the pitch lowers. Which statement is the MOST plausible question the observer might ask to explain the
phenomenon she experiences?
A. Why does the sound of the engine change as the car drives away?
B. What is the effect of the change in speed of the car and the pitch of the sound?
C. What is the relationship between the volume of the sound and the pitch of the sound
of the car?
D. How do changes in frequency and amplitude affect the pitch of the sound generated
by the car?
Physics
1 answer:
Andrei [34K]3 years ago
6 0
I believe the answer is d
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3 years ago
This problem has been solved!
lana66690 [7]

Answer:

Charge on each metal sphere will be 8\times 10^{8}C

Explanation:

We have given number of electron added to metal sphere A n=10^{12}electron

As both the spheres are connected by rod so half -half electron will be distributed on both the spheres.

So electron on both the spheres =\frac{10^{12}}{2}=5\times 10^{11}electron

We know that charge on each electron e=1.6\times 10^{-19}C

So charge on both the spheres will be equal to q=1.6\times 10^{-19}\times 5\times 10^{11}=8\times 10^{8}C

So charge on each metal sphere will be equal to 8\times 10^{8}C

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3 years ago
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4 0
3 years ago
If an atomic nucleus were the size of a dime how far away might one of its electrons be?
mihalych1998 [28]
The radius of a nucleus of hydrogen is approximately r_{n1}=1\cdot 10^{-15}m, while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: r_{e1}=5.3 \cdot 10^{-11}m

The radius of a dime is approximately r_{n2} = 9\cdot 10^{-3}m: if we assume that the radius of the nucleus is exactly this value, then we  can find how far is the electron by using the proportion
r_{n1}:r_{e1}=r_{n2}:r_{e2}
from which we find
r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.
6 0
3 years ago
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