Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
The answer is the third choice, "the brightness of the beam of light increases"
According to Einstein’s theory, an increase in the number of photons (per unit are) affects a beam of light by causing a higher intensity. <span> Sometimes, the term "brightness" is used when referring to the </span>intensity<span> of a color, a</span><span>lthough there are instances where this can be a misleading term when we try to describe </span><span>intensity</span>
A) No physical law is violated. What happens is that all this is simply impossible. On the one hand we have the "flames" released by the explosion, this can not be possible because there must be oxygen for fire to be able to burn, and in the vacuum of space there is nothing like that. On the other hand we have the "noise" that causes the explosion, this can not be because the sound needs a medium in which to move through space, on earth, the sound uses the air to transport because unlike light , this is a pressure wave. In conclusion, the only thing we could see of an explosion in the vacuum of space would be a flash of light caused by the explosion.
B) Idk what do u mean with "<span>the scene is presented from the perspective of spaceship"</span>
Answer:
Explanation:
For this case we know that the initial velocity is given by:
The final velocity on this case is given by:
And we know that it takes 8 seconds to go from 7m/s to 13m/s. We can use the following kinematic formula in order to find the acceleration during the first interval:
If we solve for the acceleration we got:
So for the other traject we assume that the acceleration is constant and the train travels for 16 s. The initial velocity on this case would be 13m/s from the first interval and we can find the final velocity with the following formula:
And if we replace we got:
