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OLEGan [10]
3 years ago
13

g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before

interacting. From this information, find the time interval required for the strong interaction to occur.
Physics
1 answer:
AnnyKZ [126]3 years ago
5 0

Answer:

Time, t=1.07\times 10^{-23}\ s

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2\times 10^{-15}\ m before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s

So, the time interval is 1.07\times 10^{-23}\ s

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From the question, specific gravity of density = 13.3

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Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

By substituting parameters, we have that

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How long does it take a microwave of power 0.2kW to sue 10000 J of energy
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Answer:

50s .

Explanation:

\frak{\pink{Given}}\begin{cases}\textsf{ The power of microvave is 0.2kW .}\\\textsf{ Amount of energy is 10000 J .}\end{cases}

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  • Rate of doing work is called power .

So from the definition , we have ;

\sf\longrightarrow Power =\dfrac{Work}{time}

  • Here the work done is equal to the energy consumed by the microwave i.e. 10000 J .So we can write it as ,

\sf\longrightarrow Power =\dfrac{Energy}{time}

\sf\longrightarrow 0.2kW = \dfrac{10^4 J }{t} \\

\sf\longrightarrow 0.2 * 1000 W =  \dfrac{10^4 J }{t}

Cross multiply ,

\sf\longrightarrow t = \dfrac{ 10^4 }{ 0.2 * 10^3}s=\dfrac{10^4}{2*10^2} s

Simplify ,

\sf\longrightarrow \boxed{\bf t = 50s}

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