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UNO [17]
4 years ago
5

A train moving near the speed of light enters a tunnel. According to a person standing in the middle of the tunnel, the back end

of the train enters the tunnel just as the front end is emerging. What happens according to someone sitting in the middle of the train?
Physics
1 answer:
Neporo4naja [7]4 years ago
4 0

Answer: The person sitting in the middle of the train sees the back of the train enter ing the tunnel before the front end comes out.

Explanation:

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Can some answer 7 a and b please
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Answer:

a.After 15 second Mr Comer's speed =  1.7 m/s

b.Distance travelled by Mr.Comer in 15 seconds  =   18.75\ m

Explanation:

a. Lets recall our first equation of motion V_{f} =V_{i} + at

Now we know that V_{i} = 0.8m/s , a = 0.06\ m/s^2 and

t = 15 \ sec

Plugging the values we have.

V_{f} =V_{i} + at

V_{f} =0.8 + 0.06 \times 15

V_{f} =0.8+ 0.9

V_{f} =1.7 m/s

Then Mr.Comer's speed after 15 sec = 1.7ms^-1

b.

Lets find the distance and recall our third equation of motion.

V_{f} ^2-V{i}^2 = 2as

So s = distance covered.

Dividing both sides with 2a we have.

\frac{V_{f} ^2-V{i}^2}{2a} = 2as

Plugging the values.

\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as

s= 18.75\ m

So Mr.Comer will travel a distance of s= 18.75\ m.

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