Since the rocket’s acceleration is 3.00 m/s^3 * t, its acceleration is increasing at the rate of 3 m/s^3 each second. The equation for its velocity at a specific time is the integral of the acceleration equation.
<span>vf = vi + 1.5 * t^2, vi = 0 </span>
<span>vf = 1.5 * 10^2 = 150 m/s </span>
This is the rocket’s velocity at 10 seconds. The equation for its height at specific time is the integral velocity equation
<span>yf = yi + 0.5 * t^3, yi = 0 </span>
<span>yf = 0.5 * 10^3 = 500 meters </span>
<span>This is the rocket’s height at 10 seconds. </span>
<span>Part B </span>
<span>What is the speed of the rocket when it is 345 m above the surface of the earth? </span>
<span>Express your answer with the appropriate units. </span>
<span>Use the equation above to determine the time. </span>
<span>345 = 0.5 * t^3 </span>
<span>t^3 = 690 </span>
<span>t = 690^⅓ </span>
<span>This is approximately 8.837 seconds. Use the following equation to determine the velocity at this time. </span>
<span>v = 1.5 * t^2 = 1.5 * (690^⅓)^2 </span>
<span>This is approximately 117 m/s. </span>
<span>The graph of height versus time is the graph of a cubic function. The graph of velocity is a parabola. The graph of acceleration versus time is line. The slope of the line is the coefficient of t. This is a very different type of problem. For the acceleration to increase, the force must be increasing. To see what this feels like slowly push the accelerator pedal of a car to the floor. Just don’t do this so long that your car is speeding!!</span>
Answer:
P(final) is 2.4 times P(initial).
Explanation:
Here we can assume that the cylinder did not break and it's volume and number of moles of gas present in the cylinder remains constant.
Given the temperature increases by a factor of 2.4. Let us assume that the initial temperature be
and the final temperature be
.
Given that 
Now we know the ideal gas equation is PV=nRT
here V=constant , n=constant , R=gas constant(which is constant).





Answer:
This equation is based on twin paradox - a phenomena where one of the twin travels to space at a speed close to speed of light and the other remains on earth. the twin from the space on return discovers that the one on earth age faster.
Solution:
= 10 years
v = 0.8c
c = speed of light in vacuum
The problem can be solved by time dilation equation:
(1)
where,
t = time observed from a different inertial frame
Now, using eqn (1), we get:

t = 16.67 years
The age of the twin on spaceship according to the one on earth = 25+16.67 =41.66 years
You should write ' 4 ' in the first box,
and ' m/s² ' in the second box.