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.
Acceleration=9.81m/s^2
initial velocity=0m/s
time=.28s
We have to find final velocity.
The equation we use is
Final velocity=initial velocity+acceleration x time
Vf=0m/s+(9.81m/s^2)(.28s)
Vf=2.7468m/s
We would round this to:
Vf (final velocity)=2.7m/s
Answer:
The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
Explanation:
Given that,
Initial velocity u= 128 ft/sec
Equation of height
....(I)
(a). We need to calculate the maximum height
Firstly we need to calculate the time
From equation (I)
Now, for maximum height
Put the value of t in equation (I)
(b). The number of seconds it takes the object to hit the ground.
We know that, when the object reaches ground the height becomes zero
Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
The correct answer is A. In the direction of applied force. This is because acceleration occurs n the direction of applied force according to Newtons second law of motion which states that the acceleration of a body is directly proportional to the applied force and takes place in the direction of force.
