Answer:
50k is the km for this is the first send
Answer:
dP/dt = 26.12 W/s
Explanation:
First, we need to find the value of dt at the instant when R₃ becomes 91.7 Ω. Therefore, we use:
dR₃/dt = 0.552 Ω/s
where,
dR₃ = Change in value of resistance 3 = 91.7 Ω - 7.42 Ω = 84.28 Ω
dt = time interval = ?
Therefore,
84.28 Ω = (0.552 Ω/s)(dt)
dt = (84.28 Ω)/(0.552 Ω/s)
dt = 152.68 s
Now, we find change in power (dP):
dP = V(R₁ + R₂ + dR₃)
dP = (42.1 V)(2.96 Ω + 7.48 Ω + 84.28 Ω)
dP = 3987.71 W
Dividing by dt:
dP/dt = 3987.71 W/152.68 s
<u>dP/dt = 26.12 W/s</u>
Answer:
544 J
Explanation:
Potential energy at top = kinetic energy at bottom + work done by friction
PE = KE + W
mgh = ½ mv² + W
W = mgh − ½ mv²
Given m = 18.0 kg, h = 3.40 m, and v = 2.48 m/s:
W = (18.0 kg) (9.8 m/s²) (3.40 m) − ½ (18.0 kg) (2.48 m/s)²
W = 544 J