Answer:
3.9 × 10^7 J
Explanation:
Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J
Solution
Since the tank is half full, the height = 2.5m
Pressure = density × gravity × height
Pressure = 900 × 9.8 × 2.5
Pressure = 22050 Pascal
The cross sectional area of the pump will be area of a circle.
A = πr^2
A = π × 15^2
A = 706.858 m^2
Using the formula
Density = mass/volume
Mass = density × volume
Mass = 900 × 706.86 × 2.5
Mass = 1590.435
Energy = mgh
Energy = 1590.435 × 9.8 × 2.5
Energy = 38965657.8 J
Since the work done = energy
Therefore, the work done = 3.9 × 10^7 J
The answer is A) variable. The depression level is a value that varies with the respondents and may inform you about the relationship between depression and other variables, such as alcohol use or sleep habits. It is also likely a discrete variable, since the answer is probably an integer (1, 2, ... 10), rather than a continuous spectrum (virtually any number between 1 and 10).
The change in the kinetic energy refers to the work done in displacing a body, thus, the change in the kinetic energy of an object refers to the work done on the object.
The correct formula to use is:
W = Initial kinetic energy - Final kinetic energy;
Where, W = change in kinetic energy
Final kinetic energy and initial kinetic energy = 1/2 MV^2
Initial velocity = 15 m/s
Final velocity = 13.5 m/s
Initial mass = 0.650 kg
Final mass = 0.950 kg
W = 1/2 [0.650* (15 *15)] - 1/2 [0.950 * (13.5 * 13.5)]
W = 146.25 - 173.13 = 26.88
Therefore, the change in kinetic energy is 26.88 J.
The negative sign has to be ignored, because change in kinetic energy can not be negative.
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