Answer:
B. the light will reach the front of the rocket at the same instant that it reaches the back of the rocket.
Explanation:
To an observer at rest in the rocket who can't see either sides of the rocket, the speed of the light is constant which means the distance to the front or the back is same and would appear to reach the rocket at the same time.
Although from the point of view of the person on the earth, the front of the rocket is travelling in opposite direction of the light while the back of the rocket is moving closer to the light. This means that the distance travelled by the light going forward will be longer going backwards. And since the speed of light is constant in both directions, the light will reach the back of the rocket before it reaches the front for the observer on the earth.
Http://www.calculator.net/pace-calculator.html?ctype=distance&ctime=05%3A00%3A00&cdistance=5&cdistanceunit=Miles&cpace=02%3A00%3A00&cpaceunit=tpm&printit=0&x=87&y=24 a pace calculator
Answer:
35, I got you bro, i got you
The way it rotates is Counter-clockwise
Answer:
Explanation:
a ) Let let the frictional force needed be F
Work done by frictional force = kinetic energy of car
F x 107 = 1/2 x 1400 x 35²
F = 8014 N
b )
maximum possible static friction
= μ mg
where μ is coefficient of static friction
= .5 x 1400 x 9.8
= 6860 N
c )
work done by friction for μ = .4
= .4 x 1400 x 9.8 x 107
= 587216 J
Initial Kinetic energy
= .5 x 1400 x 35 x 35
= 857500 J
Kinetic energy at the at of collision
= 857500 - 587216
= 270284 J
So , if v be the velocity at the time of collision
1/2 mv² = 270284
v = 19.65 m /s
d ) centripetal force required
= mv₀² / d which will be provided by frictional force
= (1400 x 35 x 35) / 107
= 16028 N
Maximum frictional force possible
= μmg
= .5 x 1400 x 9.8
= 6860 N
So this is not possible.
