A 2.31-kg rope is stretched between supports that are 10.4 m apart, and has a tension in it of 49.2 N. If one end of the rope is
1 answer:
Answer:
The resulting disturbance takes approximately 0.70 seconds to reach the other end.
Explanation:
The velocity of a disturbance on a rope depends on the physical properties of the material and the tension exerted to it, the equation is:
(1)
With v the velocity of the disturbance, T the tension of the rope and μ the linear density of the rope. The linear density of the rope is the relation between mass M and length L so we have:
(2)
Using (2) on (1):
(3)
If we assume the velocity of the disturbance is constant, it will take for the resulting disturbance to reach the other ends a time t:
(4)
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Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
