Answer:
a) P = 86720 N
b) L = 131.2983 mm
Explanation:
σ = 271 MPa = 271*10⁶ Pa
E = 119 GPa = 119*10⁹ Pa
A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²
a) P = ?
We can apply the equation
σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N
b) L₀ = 131 mm = 0.131 m
We can get ΔL applying the following formula (Hooke's Law):
ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)
⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm
Finally we obtain
L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm
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Answer:
1. Measure the temperature of the boxes and leave them unconnected.
2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.
3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.
Answer:
The Current will decrease by a factor of 2
Explanation:
Given the conditions, it should be noted that the current in the circuit is determined by the LOAD. In other words, the amount of current generator will be producing depends upon the load connected to it.
Now, as the question says, the load is reduced to half its original value, we can write:
Since, P2 = P1/2,
Dividing equations (1) and (2), we get,
P1 / (P1/2) = I1/ I2
Hence, it is proved that the current in the transmission line will decrease by a factor of 2 when load is reduced to half.
