Answer:
3.16 ×
W/
Explanation:
β(dB)=10 × 
=
W/
β=55 dB
Therefore plugging into the equation the values,
55=10
})[/tex]
5.5=
})[/tex]
= 
316227.76×
= I
I= 3.16 ×
W/
A bridge supported by vertical cables which then leads to more support from larger cables.
Answer:
2.464 cm above the water surface
Explanation:
Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.
We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3 cm^3):
weight of the block = 0.78 * 11.2^3 gr
Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.
So the weight of the volume of water displaced is:
weight of water = 1 * 11.2^2 * x
we make both weight expressions equal each other for the floating requirement:
0.78 * 11.2^3 = 11.2^2 * x
then x = 0.78 * 11.2 cm = 8.736 cm
This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:
11.2 cm - 8.736 cm = 2.464 cm
Explanation:
Value of the cross-sectional area is as follows.
A =
= 3.45 
The given data is as follows.
Allowable stress = 14,500 psi
Shear stress = 7100 psi
Now, we will calculate maximum load from allowable stress as follows.

= 
= 50025 lb
Now, maximum load from shear stress is as follows.

= 
= 48990 lb
Hence,
will be calculated as follows.

= 48990 lb
Thus, we can conclude that the maximum permissible load
is 48990 lb.
To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

Here,
h = Depth average
= Density
Moreover,

Replacing,


Finally the force


