Answer:
The comparison of unknown quantity with known quantity is called measurement
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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:
2.4s
Explanation:
The length of the pendulum = 75ft
Diameter d = 12 inches
The time period of the pendulum is given as
T = 2pi(L/g)^1/2
Then the time it takes to move from displacement to equilibrium is given as:
t = T/4
= (Pi/2)*(L/g)^1/2
= pi/2 x [(75x0.3048)/9.81]^0.5
= 1.57x[22.86/9.81)^0.5
= 2.4s
2.4 seconds is the least amount of time that it would take.
A build up of charges on a sock from a dryer
