The best answer is B. If you curve them away, they have a less tendency to be attracted to the pair of charges and would stay around the pair instead of interacting with them.
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
Answer:
Explanation:
Given that,
B(t) = B0 cos(ωt) • k
Radius r = a
Inner radius r' = a/2 and resistance R.
Current in the loop as a function of time I(t) =?
Magnetic flux is given as
Φ = BA
And the Area is given as
A = πr², where r = a/2
A = πa²/4
Then,
Φ = ¼ Bπa²
Φ(t) = ¼πa²Bo•Cos(ωt)
Then, the EMF is given as
ε(t) = -dΦ/dt
ε(t) = -¼πa²Bo • -ωSin(ωt)
ε(t) = ¼ωπa²Bo•Sin(ωt)
From ohms law,
ε = iR
Then, i = ε/R
I(t) = ¼ωπa²Bo•Sin(ωt) /R
This is the current induced in the loop.
Check attachment for better understanding
Answer:
The potential energy of the rock = 10.5 kN
Explanation:
Mass of rock = 25 kg
Acceleration due to gravity = 10 m/s²
Height = 42 m
Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity and h is the height.
PE = 25 x 10 x 42 = 10500 N = 10.5 kN
The potential energy of the rock = 10.5 kN
Answer:
The time it takes the ball to stop is 0.021 s.
Explanation:
Given;
mass of the softball, m = 720 g = 0.72 kg
velocity of the ball, v = 15.0 m/s
applied force, F = 520 N
Apply Newton's second law of motion, to determine the time it takes the ball to stop;
Therefore, the time it takes the ball to stop is 0.021 s.
