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kakasveta [241]
3 years ago
9

Can a ratio be expressed as a percentage

Physics
1 answer:
cupoosta [38]3 years ago
7 0
Yes, ratio can be expressed in percentage.
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What is the mass of an object moving with 80N of force and an acceleration of 8 m/s2?
OverLord2011 [107]

Answer:

10 Kg

Explanation:

Force is equal to mass times acceleration

therefore mass is equal to force divided by acceleration

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3 years ago
What length of tube would be required to produce a second tone under the same experimental conditions? Explain your answer.
Alisiya [41]

There is an indirect relationship between length and frequency. The longer the length the pipe has, the higher frequency it is. The shorter the length the pipe has, the lower frequency it is.

<u>Explanation:</u>

The four properties of the string that affect its frequency are length, diameter, tension, and density. These properties are described below: When the length of a string is changed, it will vibrate with a different frequency. Shorter strings have higher frequency and therefore higher pitch.

The longer the tube is the lower the pitch of the note that it can emit. When a tube is heated it expands and so is longer! As the gas in the tube gets warmer the molecules move faster, that means they can carry the vibrations of the sound wave more rapidly and so the pitch goes up.

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3 years ago
How would clearing a forest to plant corn affect an enviroment
Oksanka [162]
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4 0
3 years ago
An empty 230 kg elevator accelerates upward
elena-s [515]

Answer:

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3 0
3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
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