3_3. A pump is required to lift a mass of 800 kg of water per minute from a well
ID: 2017666 • Letter: 3
Question
3_3. A pump is required to lift a mass of 800 kg of water per minute from a well of depth 14.7 m and eject it with a speed of 17.4 m/s.a. How much work is done per minute in lifting the water?
b. How much in giving the water the kinetic energy it has when ejected?
K =?
c. What must be the power output of the pump?
P =?
3_4. The Grand Coulee Dam is 1270 m long and 170 m high. The electrical power output from generators at its base is approximately 1940 MW.
a.How many cubic meters of water must flow from the top of the dam per second to produce this amount of power if 95.0 % of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000 .)
3_5. A small rock with mass 0.12 kgis fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 degrees with the vertical. Air resistance is negligible.
a. What is the speed of the rock when the string passes through the vertical position?
Express your answer using two significant figures.
v =?
b.What is the tension in the string when it makes an angle of 45 degrees with the vertical? Express your answer using two significant figures.
c. What is the tension in the string as it passes through the vertical? Express your answer using two significant figures.
Explanation / Answer
Given Mass to be lifted , m = 800 kg / min Depth of the well , h = 14.7 m Speed of the water , v = 17.4 m/s a) Work done per minute , W = mgh = 800kg / min * 9.8 m/s^2 *14.7 m W = 1.15 *10^5 J per minute b) Kinetic energy, KE = 1/2 mv^2 = 0.5 *800 kg *(17.4 m/s)^2 = 1.21 *10^5 J c) Power , P = W / t = (1.15 *10^5) / 60 s = 1916.67 W * Post only one question *Related Questions
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