It’s light seconds, now don’t mistake it as time
Complete question is;
A copper wire has a diameter of 4.00 × 10^(-2) inches and is originally 10.0 ft long. What is the greatest load that can be supported by this wire without exceeding its elastic limit? Use the value of 2.30 × 10⁴ lb/in² for the elastic limit of copper.
Answer:
F_max = 28.9 lbf
Explanation:
Elastic limit is simply the maximum amount of stress that can be applied to the wire before it permanently deform.
Thus;
Elastic limit = Max stress
Formula for max stress is;
Max stress = F_max/A
Thus;
Elastic limit = F_max/A
F_max is maximum load
A is area = πr²
We have diameter; d = 4 × 10^(-2) inches = 0.04 in
Radius; r = d/2 = 0.04/2 = 0.02
Plugging in the relevant values into the elastic limit equation, we have;
2.30 × 10⁴ = F_max/(π × 0.02²)
F_max = 2.30 × 10⁴ × (π × 0.02²)
F_max = 28.9 lbf
Answer:
Efficiency = 10.2 %
Explanation:
Given the following data;
Mass = 70 kg
Height = 50 m
Velocity = 10 m/s
We know that acceleration due to gravity is equal to 9.8 m/s².
To find the efficiency of energy conversion from potential to kinetic;
First of all, we would determine the potential energy;
P.E = mgh
P.E = 70 * 9.8 * 50
P.E = 34300 J
For the kinetic energy;
K.E = ½mv²
K.E = ½ * 70 * 10²
K.E = 35 * 100
K.E = 3500
Therefore, Input energy, I = 34300 J
Output energy, O = 3500 J
Next, we find the efficiency;
Efficiency = O/I * 100
Substituting into the formula, we have;
Efficiency = 3500/34300 * 100
Efficiency = 0.1020 * 100
Efficiency = 10.2 %
Answer:
The machine used is called a squaring shear, power shear, or guillotine.
Explanation:
I would think it would be the same if you are weighting the dish and the Ice cube at the same time anyway. Not to sure though I'm a beginner and math is complicated for me sometimes.
