⭐Hey
⭐Here we can use the formula for the body that is in free fall state. .
⭐that is
↪V = underoot 2*g*h
↪So the velocity will be ...
↪V = underoot 2 * 10 * 100
↪V = underoot 2000
↪V = 20 *underoot 5
↪V = 20 ( 2.23)
↪V = 44.6 m/s
Simile, as the example compares the person with a cheetah using “as” (a simile uses “like” or “as” for comparisons between two things).
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>
The hot discharge gas from the refrigerant compressor is normally cooled and condensed at high pressure. This is then passed through an 'Expansion' valve which decreases the pressure to a low level causing expansion of the refrigerant liquid.
<span>The liquid partially vapourises causing a 'Joule's/Thompson' refrigeration effect' which decreases temperature of the refrigerant which then passes to an evaporator coil in the air circulation system of the building. </span>
<span>In the evaporator coil, the heat exchange between the cold refrigerant and the warm air of the building, vaporises and heats the refrigerant which returns to the compressor. </span>
<span>The cycle is repeated until the air temperature reaches the thermostat set-point and switches off the system. </span>
<span>As a Heat pump, the hot refrigerant gas is not evaporating and condensing. </span>
<span>From the compressor discharge, the hot gas is by-passing the cooler/condenser unit and the expansion valve and passes directly to the 'evaporator' coils but now, as the heating medium for the air circulation system where it's cooled by the heat exchange between the hot gas and the cooler air in the building and returns to the compressor in a continuous cycle. </span>
<span>A Thermostat in the system starts and stops the compressor motor according to the heat or cool temperature settings.</span>
Answer:
The change in momentum of the ball is 24 kg-m/s
Explanation:
It is given that,
Mass of the ball, m = 1 kg
Initial velocity of the ball, u = -12 m/s (in downwards)
Final velocity of the ball, v = +12 m/s (in upward)
We need to find the change in momentum of the ball.
Initial momentum of the ball, 
Final momentum of the ball, 
Change in momentum of the ball, 
So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.
