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GrogVix [38]
3 years ago
5

A super train is moving along a track at a speed close to the speed of light. you are watching the train from the ground. you ob

serve lightning to strike in two places along the track, a mile apart, at precisely the same time. what would someone on the train say?
Physics
1 answer:
ryzh [129]3 years ago
7 0
The people in the train will say that it super fast, because the speed of light is about 186,411 mile / s, so a mile will only took 0.0000054 s to travel it. so it is super fast and i think the people boarding the train might suffer physically due to intensity of the force applied to their body
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Which of these was a characteristic of the Rutherford model of the atom, but not the Thomson model? (2 points)
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Thomsons model( commonly the plum pudding model) - can also be described by a watermelon cut horizontally. The red of the water melon is a positively charged cloud with the seeds representing the negatively charged electrons sprinkled through out the positive cloud. 

Rutherford had a model with a dense positively charged nucleus at the center. This nucleus was encircled by negatively charged atoms.

Answer: <span>b. positive charge concentrated at the center of the atom</span>
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Answer:

an explanation on how the natural world works

Explanation:

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What two quantities affect the force of attraction between two charges
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A beam of protons moves in a circle of radius 0.20 m. The protons move perpendicular to a 0.36-T magnetic field. (a) What is the
gregori [183]

Answer:

a) v=6.898\times 10^{6}\ m.s^{-1} is the speed of each proton

b) F_c=3.97\times 10^{-13}\ N

Explanation:

Given:

radius of path of motion, r=0.2\ m

we know charge on protons, q=1.6\times 10^{-19}\ C

magnetic field strength, B=0.36\ T

we've mass of proton, m=1.67\times 10^{-27}\ kg

a)

From the equivalence of magnetic force and the centripetal force on the proton:

F_B=F_C

q.v.B=\frac{m.v^2}{r}

q.B=\frac{m.v}{r}

where:

v = speed of the proton

(1.6\times 10^{-19})\times 0.36=\frac{1.67\times 10^{-27}\times v}{0.2}

v=6.898\times 10^{6}\ m.s^{-1} is the speed of each proton

b)

Now the centripetal force on each proton:

F_c=m.\frac{v^2}{r}

F_c=1.67\times 10^{-27}\times \frac{(6.898\times 10^6)^2}{0.2}

F_c=3.97\times 10^{-13}\ N

6 0
4 years ago
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dlinn [17]

Answer:

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