Answer:
189.15cy
Explanation:
To understand this problem we need to understand as well the form.
It is clear that there is four wall, two short and two long.
The two long are 
The two long are 
The two shors are 
The height and the thickness are 14ft and 0.83ft respectively.
So we only calculate the Quantity of concrete,
![Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3](https://tex.z-dn.net/?f=Q_c%20%3D%20%5B%282%2A122.08%29%2B%282%2A86-375%29%5D%2A14%2A0.833%5C%5CQ_c%3D4864.02ft%5E3)
That in cubic yards is equal to 
Hence, we need order 5% plus that represent with the quantity

Answer:
Explanation:
volume of 20.9 N
= 20.9 / 11.5 m³
= 1.8174 m³
In one hour 1.8174 m³ flows
in one second volume flowing = 1.8174 / 60 x 60
= 5 x 10⁻⁴ m³
Rate of volume flow = 5 x 10⁻⁴ m³ / s .
Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ =
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ =
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
Answer:
The FSM uses the states along with the generation at the P output on each of the positive edges of the CLK. The memory stores the previous state in the machine and the decoder generates a P output based on the previous state.
Explanation:
The code is in the image.