2 m^3 of an ideal gas is compressed from 100 kPa to 200 kPa. As a result of the process, the internal energy of the gas increases by 10 kJ, and 140 kJ of heat is lost to the surroundings. What is the work done by the gas during the process?
The information we have is that as a result of the process 140 kJ of heat is lost, so Q=Q_out. And the internal energy of the gas increases by 10 kJ so ∆E = 10 kJ. Using this values we can find the work: Q_out-W=∆E -140 [kJ] - W= 10 [kJ] -140 [kJ] - 10 [kJ] = W W= - 150 [kJ] The negative sign indicates that the work was done on the gas which is correct because the gas was compressed.
A train of mass 200 KN has a frictional resistance of 5 N per KN. The speed of the train, at the top of an incline of 1 in 80 is 45 Km/hr. Find the speed of the train after running down the incline for 1 Km.
Given data, Mass m = 200 KN, Frictional resistance Fr= 5 N/KN, sinθ= 1/80 = 0.0125, Initial velocity u = 45 Km/hr = 12.5 m/sec, s = 1 km = 1000 m Total frictional resistance = 5 X 200 = 1000 N = 1 KN Force responsible for sliding = Wsinθ = 200 X 0.0125 = 2.5 KN Now, Net force, F = F – Fr = ma 2.5 – 1= (200/9.81)a a = 0.0735 m/sec^2 Apply the equation, v^2= u^2 + 2as v^2 = 0 + 2 X 0.0735 X 1000 Answer: v= 12.1 m/sec
What is belt? How many types of belt are used for power transmission?
The power or rotary motion from one shaft to another at a considerable distance is usually transmitted by means of flat belts, Ve belts or ropes, running over the pulley. But the pulleys contain some friction. Three types of belt are used for power transmissions: Flat belt, V- belt and Circular Belt