Answer:
(B) The total internal energy of the helium is 4888.6 Joules
(C) The total work done by the helium is 2959.25 Joules
(D) The final volume of the helium is 0.066 cubic meter
Explanation:
(B) ∆U = P(V2 - V1)
From ideal gas equation, PV = nRT
T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3
P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal
∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules
(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal
Assuming a closed system
(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules
(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter
We know that the Delta E + W(Work done by non-conservative
forces) = 0 (change of energy)
In here, the non-conservative force is the friction force
where f = uN (u =kinetic friction coefficient)
W= f x d = uNd ; N=mg
Delta E = 1/2 mV^2 -1/2mVi^2
umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term)
This will then give us:
1/2Vi^2-ugd = 1/2V^2
V^2 = Vi^2 - 2ugd
So plugging in our values, will give us:
V= Sqrt (5.6^2 -2.3^2)
=sqrt (26.07)
= 5.11 m/s
Answer:
golf all on grass covered hill in sunlight
Explanation:
someone pouring/spinning top has a lot of motion going on. a roller coaster has force acting on it which increases the motion.
At position of maximum height we know that the vertical component of its velocity will become zero
so the object will have only horizontal component of velocity
so at that instant the motion of object is along x direction
while if we check the acceleration of object then it is due to gravity
so the acceleration of object is vertically downwards
so it is along y axis
so here these two physical quantities are perpendicular to each other
so correct answer would be
<em>C)At the maximum height, the velocity and acceleration vectors are perpendicular to each other. </em>
Answer:
Explanation:
Diameter of pool = 12 m
radius of pool, r = 6 m
Total height raised, h = 3 + 2.5 = 5.5 m
density of water, d = 1000 kg/m³
Mass of water, m = Volume of water x density
m = πr²h x d
m = 3.14 x 6 x 6 x 5.5 x 1000
m = 113040 kg
Work = m x g x h
W = 113040 x 9.8 x 5.5
W = 6092856 J
