Answer:
The strain is 0.0947.
Explanation:
Strain is the ration of change in dimension to the original dimension under application of load. Depending upon load, the bar may stretch or compress. So, the strain can be negative in case of compression or positive in case of tension.
Step1
Given:
Original length of the steel bar is 0.66 ft or 7.92 in.
Stretch in the bar is of 0.75 in.
Calculation:
Step2
Strain is calculated as follows:


e = 0.0947
Positive sign is for stretching condition of the bar.
Thus, the strain is 0.0947.
Answer:
Proof is as follows
Proof:
Given that , 
<u>for any function f with period T, RMS is given by</u>
<u />
<u />
In our case, function is 
![RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac} + V_{dc}]^{2} } \, dt }](https://tex.z-dn.net/?f=RMS%20%3D%20%5Csqrt%7B%5Cfrac%7B1%7D%7BT%7D%5Cint%5Climits%5ET_0%20%7B%5BV_%7Bac%7D%20%2B%20V_%7Bdc%7D%5D%5E%7B2%7D%20%7D%20%5C%2C%20dt%20%20%7D)
Now open the square term as follows
![RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac}^{2} + V_{dc}^{2} + 2V_{dc}V_{ac}] } \, dt }](https://tex.z-dn.net/?f=RMS%20%3D%20%5Csqrt%7B%5Cfrac%7B1%7D%7BT%7D%5Cint%5Climits%5ET_0%20%7B%5BV_%7Bac%7D%5E%7B2%7D%20%2B%20V_%7Bdc%7D%5E%7B2%7D%20%2B%202V_%7Bdc%7DV_%7Bac%7D%5D%20%7D%20%5C%2C%20dt%20%20%7D)
Rearranging terms

You can see that
- second term is square of RMS value of Vac
- Third terms is average of VdcVac and given is that average of

so
![RMS = \sqrt{\frac{1}{T}TV_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }](https://tex.z-dn.net/?f=RMS%20%3D%20%5Csqrt%7B%5Cfrac%7B1%7D%7BT%7DTV_%7Bdc%7D%5E%7B2%7D%20%20%20%2B%20%5BRMS~~%20of~~%20V_%7Bac%7D%5D%5E2%20%7D)
![RMS = \sqrt{V_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }](https://tex.z-dn.net/?f=RMS%20%3D%20%5Csqrt%7BV_%7Bdc%7D%5E%7B2%7D%20%20%20%2B%20%5BRMS~~%20of~~%20V_%7Bac%7D%5D%5E2%20%7D)
So it has been proved that given expression for root mean square (RMS) is valid
Answer:
2.65 MPa
Explanation:
To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

<u>Where:</u>
p: is the gage pressure = 108 kPa
r: is the inner radius of the ball
t: is the thickness = 3 mm
Hence, we need to find r, as follows:

<u>Where:</u>
d: is the outer diameter = 300 mm

Now, we can find the normal stress (σ) in the wall of the basketball:
Therefore, the normal stress is 2.65 MPa.
I hope it helps you!