Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
Answer:
Explanation:
Given that,
The mutual inductance of the two coils is
M = 300mH = 300 × 10^-3 H
M = 0.3 H
Current increase in the coil from 2.8A to 10A
∆I = I_2 - I_1 = 10 - 2.8
∆I = 7.2 A
Within the time 300ms
t = 300ms = 300 × 10^-3
t = 0.3s
Second Coil resistance
R_2 = 0.4 ohms
We want to find the current in the second coil,
The same induced EMF is in both coils, so let find the EMF,
From faradays law
ε = Mdi/dt
ε = M•∆I / ∆t
ε = 0.3 × 7.2 / 0.3
ε = 7.2 Volts
Now, this is the voltage across both coils,
Applying ohms law to the second coil, V=IR
ε = I_2•R_2
0.72 = I_2 • 0.4
I_2 = 0.72 / 0.4
I_2 = 1.8 Amps
The current in the second coil is 1.8A
Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
Answer:
Temperature : 92.9 F
Internal Energy change: -2.53 Btu/lbm
Explanation:
As
mh1=mh2
h1=h2
In table A-11 through 13E
p2=120Psi, h1= 41.79 Btu/lbm,
u1=41.49
So T1=90.49 F
P2=20Psi
h2=h1= 41.79 Btu/lbm
T2= -2.43F
u2= 38.96 Btu/lbm
T2-T1 = 92.9 F
u2-u1 = -2.53 Btu/lbm
