Answer:
Explanation:
From the given information:
Strain fracture toughness 
= 75 MPa
Tensile stress 
= 361 MPa
Value of Y = 1.03
Thus, the minimum length of the critical interior surface crack which will result to fracture can be determined by using the formula:
![a_c = \dfrac{1}{\pi} ( \dfrac{k_k}{\sigma Y})^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ \dfrac{75 \times \sqrt{10^3}}{361 \times 1.03 } \Big]^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ 6.378474693\Big]^2 \\ \\ \mathbf{ a_c = 12.95 \ mm}](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cdfrac%7B1%7D%7B%5Cpi%7D%20%28%20%5Cdfrac%7Bk_k%7D%7B%5Csigma%20Y%7D%29%5E2%20%5C%5C%20%5C%5C%20%20a_c%20%3D%20%5Cdfrac%7B1%7D%7B%5Cpi%7D%20%5CBig%20%5B%20%5Cdfrac%7B75%20%5Ctimes%20%5Csqrt%7B10%5E3%7D%7D%7B361%20%5Ctimes%201.03%20%7D%20%5CBig%5D%5E2%20%5C%5C%20%5C%5C%20%20a_c%20%3D%20%5Cdfrac%7B1%7D%7B%5Cpi%7D%20%5CBig%20%5B%206.378474693%5CBig%5D%5E2%20%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%20a_c%20%3D%2012.95%20%5C%20mm%7D)
Answer:
331809.5gallon/hr or 92.16gallon/s
Explanation:
What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C
convert 9.4 acre to inches we have=5.896*10^7
How to calculate Peak runoff discharge
1. take the dimension of the roof
2. multiply the dimension by the n umber of inches of rainfall
3. Divide by 231 to get gallon equivalence (because 1 gallon = 231 cubic inches)
5.896*10^7*1.3
7.66*10^7 cubic inches/hr
1 gallon=231 cubic inches
7.66*10^7 cubic inches=331809.5gallon/hr or 92.16gallon/s
this is gotten by converting 1 hr to seconds
331809.5gallon/hr /3600s=92.16gallon/s
Answer:
<em>Python code is as follows:
</em>
********************************************************************************
#function to get number up to any number of decimal places
def toFixed(value, digits):
return "%.*f" % (digits, value)
print("Enter the price: ", end='', flush=True) #prompt for the input of price
price = float(input()) #taken input
totalCost = price + 0.05 * price #calculating cost
print("Total Cost after the sales tax of 5% is applied: " + toFixed(totalCost,2)) #print and output totalCost
************************************************************************************
Answer:
2m
Explanation:
The modulus of rigidity was omitted in the question and I'm using the modulus of rigidity of structural steel (G=79 GPa)
For the shaft, the maximum moment
T=π×rho×d³/16
Where π=3.14
Rho= torsional stress=109MPa
d=diameter of of shaft= 21mm
Inserting values of π, rho and d
T= 198,230.54Nmm
Polar moment of the shaft is given by
J= π D⁴/ 32
Also torsional moment (in radian) of the shaft is given as
α = L ×T / (J× G)
α = 32× L×T / (G× π× D⁴)
And in degrees,
αdegrees = 584× L×T / (G×D⁴)
Where L=length of shaft and G is modulus of rigidity
L=α×d⁴×G/584×T
Where α=15⁰, d=0.021m, G=79×10⁹Pa, T=198.23Nm
Substituting values for α, d, G and T
L=1.99m approximately 2m
