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
119.7kPa
Explanation:
Given parameters:
Pressure of gas in balloon = 111kPa
Temperature of gas = 22°C
Final temperature = 45°C
Unknown:
Final pressure = ?
Solution:
Since the gases in the balloon have the same number of moles. We can apply a derivative of the combined gas law to solve this problem.
At constant volume the pressure of a given mass of gas varies directly with the absolute temperature.

P1 is the initial pressure
P2 is the final pressure
T1 is the initial temperature
T2 is the final temperature
convert from celcius to kelvin:
tK = 273 + tC
T1 = 273 + 22 = 295K
T2 = 273 + 45 = 318K

P2 = 119.7kPa
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Ideal gas brainly.com/question/13064292
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Answer:
A magnet is demagnetized. There is a decrease in chemical potential energy. There is an increase in thermal energy. And there is an increase in energy in the form of demagnetization.
i believe thaT IT IS B BUT PLEASE TELL ME IF IM WRONG I WAS JUST USING LOGIC
Answer:
The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1
Explanation:
Let the weight that is hooked to two springs be w.
Spring#1:
Force constant= k
let x1 be the extension in spring#1
Therefore by balancing the forces, we get
Spring force= weight
⇒k·x1=w
⇒x1=w/k
Energy stored in a spring is given by
where k is the force constant and x is the extension in spring.
Therefore Energy stored in spring#1 is, 
⇒
⇒
Spring #2:
Force constant= 2k
let x2 be the extension in spring#2
Therefore by balancing the forces, we get
Spring force= weight
⇒2k·x2=w
⇒x2=w/2k
Therefore Energy stored in spring#2 is, 
⇒
⇒
∴The ratio of the energy stored by spring #1 to that stored by spring #2 is
2:1