Answer:
The distance the log has moved by the time Ernie reaches Bur is 1.33 m.
Explanation:
give information:
The log is 3.0 m long and has mass 20.0 kg.
Burt has mass 30.0 kg; Ernie has mass 40.0 kg
Ernie has mass 40.0 kg.
to find the distance, first, we have to calculate the center of mass
X = ∑ m x /∑m
= (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)
= 150/90
= 5/3
when Ernie walk, the center of the mass is
X = (70 x 0) + (20 x (3/2))/(70 + 20)
= 30/90
= 1/3
the distance of log moved = 5/3 - 4/3 = 1.33 m
Answer:
Z = R, i = V/Z, w = √1 / LC
Explanation:
In an RLC circuit the impedance of the circuit is
Z = √[R² + (
)²
Where
= wL
X_{L} = 1 / wC
They are the reactances of the inductor and the capacitor, in this case the current advances to the voltage in the first and is delayed from the voltage in the second, so when the two values give the same reactance the current goes in phase with the voltage and the impedance is minimal
Z = R
V= i Z
i = V/Z
Therefore the current is maximum, this occurs when
w = √1 / LC
Saying that this is the resonant frequency
The Splitting of a nucleus
so, question number 10 answer is 82 watts
Answer:
the rate of heat transfer after the system achieves steady state is -3.36 kW
Explanation:
Given the data in the question;
mass of water m = 50 kg
N = 300 rpm
Torque T = 0.1 kNm
V = 110 V
I = 2 A
Electric work supplied W₁ = PV = 2 × 110 = 220 W = 0.22 kW
Now, work supplied by paddle wheel W₂ is;
W₂ = 2πNT/60
W₂ = (2π × 0.1 × 300) / 60
W₂ = 188.495559 / 60
W₂ = 3.14 kW
So the total work will be;
W = 0.22 + 3.14
W = 3.36 kW
Hence total work done on the system is 3.36 kW.
At steady state, the properties of the system does not change so the heat transfer will be 3.36 KW.
The heat will be rejected by the system so the sign of heat will be negative.
i.e Q = -3.36 kW
Therefore, the rate of heat transfer after the system achieves steady state is -3.36 kW
