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g100num [7]
3 years ago
11

Suppose an atom has a mass number of 23. What does this mean?

Physics
2 answers:
lawyer [7]3 years ago
6 0
The number of protons in the nucleus dose not equal the number of neutrons.
melisa1 [442]3 years ago
5 0
Vandanium, that's what 23 on the periodic table is.
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An object will not change its motion unless an outside force acts on it true or false
Gre4nikov [31]

Answer:

True!

Explanation:

Newton's First Law :)

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What is the speed of sound for a noise that travels 2km in 5.8s?​
Gnesinka [82]

Answer:

Explanation:

Speed is defined as the rate at which an object covers a particular distance. So the formula for determining speed is given as the ratio of distance to time taken for covering that distance.

Speed = Distance/Time

As here the distance is given in km units and time in s units, so the units of any one parameter should be changed. Since we know that speed of sound is always about 300 m/s. So it is better to convert the unit of distance from km to m.

Hence, now the distance traveled by the noise is 2000 m and time taken is 5.8 s.

So the speed of noise = Distance/Time = 2000/5.8=345 m/s.

Thus, the speed of noise is slightly greater than the speed of sound and it is found to be 345 m/s.

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4 years ago
If you throw a raw egg against a wall, you'll break it, but if you throw it with the same speed into a sagging sheet it won't br
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A scientist heats up a container full of neon gas. What will most likely happen to the container as the gas heats up?
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3 years ago
Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o
o-na [289]

Answer:

<em>20.08 Volts</em>

Explanation:

<u>Parallel Connection of Capacitors</u>

The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is

\displaystyle V_1=\frac{Q_1}{C_1}

\displaystyle V_2=\frac{Q_2}{C_2}

They are both the same after connecting them, thus

\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}

Or, equivalently

\displaystyle Q_2=\frac{C_2Q_1}{C_1}

The total charge of both capacitors is

\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C

Now we compute Q1 from the equation above

\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C

The final voltage of any of the capacitors is

\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V

7 0
3 years ago
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