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:
more than 90%
Explanation:
In the unknown liquid the buoyant force and weight relation is

It can be seen that if the density decreases the buoyant force decreases.
If the object is already 90% submerged in water then, if the other liquid has density less than that of water the object will be submerged more than 90%.
Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma

Let d is the distance moved in 2.25 s. Using second equation of motion,

So, it will move 6.32 m from rest in 2.25 seconds.
Answer:
The value is 
Explanation:
From the question we are told that
The velocity which the rover is suppose to land with is
The mass of the rover and the parachute is
The drag coefficient is
The atmospheric density of Earth is 
The acceleration due to gravity in Mars is 
Generally the Mars atmosphere density is mathematically represented as

=> 
=> 
Generally the drag force on the rover and the parachute is mathematically represented as

=>
=>
Gnerally this drag force is mathematically represented as

Here A is the frontal area
So

=> 
=> 