<span>137200000 watts
or 137200 kilowatts
The formula for power is
P= dhrg
Where
P = Power in watts
d = density of water (~1000 kg/m^3)
h = height in meters
r = flow rate in cubic meters per second,
g = acceleration due to gravity of 9.8 m/s^2,
Plugging in the known values, we get
P = 1000 kg/m^3 * 80 m * 175 m^3/s * 9.8 m/s^2
P = 80000 kg/m^2 * 175 m^3/s * 9.8 m/s^2
P = 14000000 kg m/s * 9.8 m/s^2
P = 137200000 kg m^2/s^3
P = 137200000 watts
or 137200 kilowatts
The above figure assumes 100% efficiency which is impossible. A good efficiency would be 90% so the actual power available would be close to 0.90 * 137200 = 123480 kilowatts</span>
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:
0.45 miles
Explanation:
since, car will be running on the edge of track, distance covered in one lap by car will be equal to circumference of track.\
we will use value of pi as 22/7
Circumference for any circular shape is given by 
Hence, in one lap distance will be covered
in 7 laps 
distance will be covered.
but is given that 7 laps should be equal to 10 miles
so 10 miles will be equal to 
Thus, diameter of track will be 0.45 miles to meet the requirement.
- P is power
- R is resistance
Hence
- Therefore if power is low then resistance will be high.
The first bulb has less power hence it has greater filament resistance.
