A supertrain with a proper length of 100 m travels at a speed of 0.950c as it passes through a tunnel having a proper length of 50.0 m. As seen by a trackside observer, is the train ever completely within the tunnel? If so, by how much do the train’s ends clear the ends of the tunnel?

19m

Explanation:

we have proper length L = 100m

the speed of the train v = 0.95

the speed of light is given as = 3x10⁸

length of the tunnel is given as = 50 meters

we can solve for the lenght contraction as

LX√1-v²/c²

= 100 * √1-(0.95*3x10⁸)²/(3x10⁸ )

= 31.22 metres

the train would be well seen at

50 - 31.22

= 18.78

= this is approximately 19 metres

we conclude tht the trains ends clears the ends of the tunnel by 19 meters.

thank you!

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form

not too sure thinking all accelerations are pointed down