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kifflom [539]
3 years ago
10

Which one of the following statements does not accurately describe vibrations?

Physics
1 answer:
Paladinen [302]3 years ago
4 0
A. Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.
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What is an electric circuit?
klemol [59]

<h3>Answer: any path that allows electrons to flow</h3>

An electrical circuit is a path in which electrons from a voltage or current source flow. ... The part of an electrical circuit that is between the electrons' starting point and the point where they return to the source is called an electrical circuit's "load".

7 0
3 years ago
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The potential energy of a negative charge moved from point A to point B will increase.A negative charge moved from point A to po
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Answer:

<em>The K.E from A to B won't increase...</em>

Explanation:

That's because the P.E from A to B is increasing. The K.E will increase if charge moves from a higher potential to a lower potential i.e., from B to A.

That is the reason there is no effect on net K.E when moving from a potential to same potential over and over (A to C).

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3 years ago
An isolated parallel-plate capacitor has a surface charge density. If the space between the plates is filled with a material of
anygoal [31]

Answer:

Explanation:

2\sqrt{34}

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2 years ago
Question 25 of 30
VARVARA [1.3K]

Answer:

it's B. circuit a and b are series circuit while c is parallel

7 0
2 years ago
The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

M_e = mass of earth

Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

r_2 = 1 AU

T_2 = 1 year

M_s = mass of Sun

now we have

\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}

12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

4 0
3 years ago
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