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Lesechka [4]
3 years ago
10

Earth’s atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed th

rough the atmosphere, each square meter of Earth’s surface would intercept protons at the average rate of 1500 protons per second.What would be the electric current intercepted by the total surface area of the planet?
Physics
1 answer:
gulaghasi [49]3 years ago
4 0

Answer:

The current would be 0.12A

Explanation:

Well, the formula for electric current is

i=\frac{q}{t}

where i is current, q is charge and t is time. Then we have to calculate the charge. If we know that the charge of a proton is 1.6*10^{-19} C, then we calculate the charge of 1500 protons as it follows:

q=1500*1.6*10^{-19} C\\q=2.4*10^{-16} C

So the current for each square meter of Earth's surface is

i=\frac{2.4*10^{-16} C}{1s} \\i=2.4*10^{-16} A

Finally, knowing the surface of the Earth is 510065284702000 m^{2}, we can calculate the total current i_{t}

i_{t}= 2.4*10^{-16} A*510065284702000\\i_{t}=0.12A

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4 years ago
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Professor Stefanovic is spinning a bucket of water by extending his arm and rotating his shoulder in class to show the effects o
anyanavicka [17]

Answer:

a ) 2.368 rad/s

b) 3.617 rad/s

Explanation:

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i.e

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where;

F_{inward} = m *a_c

F_{inward} = m*r* \omega^2

Also

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F_T = 0          since; that is the initial minimum angular velocity to keep the water in the bucket

Now; we can rewrite our equation as :

mr \omega^2= mg + 0\\\omega^2 = \frac{m*g}{m*r}\\\omega^2 = \frac{g}{r}\\\omega = \sqrt{\frac{g}{r} \ \  }     ------ equation \ \ \ {2}

So; Given that:

The rope that is attached to the bucket is lm long  and his arm is 75 cm long.

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\omega = \sqrt{ \frac{9.81}{1.75}}\\\omega = 2.368 \  rad/s

b) if he detaches the rope and spins the bucket by holding it with his hand ; then the radius = 0.75 m

∴

\omega = \sqrt{ \frac{9.81}{0.75}}\\\omega = 3.617 \  rad/s

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4 years ago
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Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li
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So to solve this problem, we must convert units and then cancel units simultaneously:

(\frac{982,080,000ft}{s})*(\frac{32,000,000s}{1})*(\frac{12inches}{1ft})=3.77x10^{17} inches

Therefore we know that the distance that light travels in one year is 3.77x10^{17} inches.

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