Answer:
![\boxed {\boxed {\sf 6 \ meters}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%206%20%5C%20meters%7D%7D)
Explanation:
Work is the product of force and distance.
![W=F*d](https://tex.z-dn.net/?f=W%3DF%2Ad)
We know that 96 Joules of work were done and a 16 Newton force was applied to the object.
Substitute the values into the formula.
![96 \ J= 16 \ N * d](https://tex.z-dn.net/?f=96%20%5C%20%20J%3D%2016%20%5C%20N%20%2A%20d)
First, let's convert the units. This will make cancelling units easier later in the problem. 1 Joule (J) is equal to 1 Newton meter (N*m), so the work of 96 Joules equals 96 Newton meters.
![96 \ N*m= 16 \ N * d](https://tex.z-dn.net/?f=96%20%5C%20N%2Am%3D%2016%20%5C%20N%20%2A%20d)
Now, solve for distance by isolating the variable, d. It is being multiplied by 16 Newtons and the inverse of multiplication is division. Divide both sides of the equation by 16 N.
![\frac {96 \ N*m}{16 \ N}= \frac{16 \ N *d}{16 \ N}](https://tex.z-dn.net/?f=%5Cfrac%20%7B96%20%5C%20N%2Am%7D%7B16%20%5C%20N%7D%3D%20%5Cfrac%7B16%20%5C%20N%20%2Ad%7D%7B16%20%5C%20N%7D)
![\frac {96 \ N*m}{16 \ N}=d](https://tex.z-dn.net/?f=%5Cfrac%20%7B96%20%5C%20N%2Am%7D%7B16%20%5C%20N%7D%3Dd)
The units of Newtons cancel.
![\frac {96}{16} \ m = d](https://tex.z-dn.net/?f=%5Cfrac%20%7B96%7D%7B16%7D%20%5C%20m%20%3D%20d)
![6 \ m = d](https://tex.z-dn.net/?f=6%20%5C%20m%20%3D%20d)
The object moved a distance of <u>6 meters.</u>
The loudness or sound intensity of the roaring truck engine is inversely proportional to the square of the distance. it follows the law of squares that is
r1^2 / r2^2 = L2/L1
8^2 / 4^2 = L2 / 100 dB
L2 = 400 dB
The answer to this problem is 400 dB increased frim 100 dB.
Ok then what’s your question?i will help u if u want
Answer:
The submerged effective density is 86.93 kN/m³ or 8.693 Mg/m³
Explanation:
Given;
wet density of soil sample = 2.5 Mg/m³ = 25 kN/m³
Specific gravity of solid particle = 2.7
The dry unit weight of soil;
![\gamma _d = \frac{\gamma _t}{1 +w} = \frac{25}{1+0.25} = 20 \ kN/m^3](https://tex.z-dn.net/?f=%5Cgamma%20_d%20%3D%20%5Cfrac%7B%5Cgamma%20_t%7D%7B1%20%2Bw%7D%20%3D%20%5Cfrac%7B25%7D%7B1%2B0.25%7D%20%3D%2020%20%5C%20kN%2Fm%5E3)
for undisturbed state, the volume of the soil is;
![V_s =\frac{\gamma _d}{G_s \gamma _w} = \frac{20}{2.7*9.81} = 0.76 \ m^3\\\\Void \ volume, V_v = 1-0.76 = 0.24 \ m^3](https://tex.z-dn.net/?f=V_s%20%3D%5Cfrac%7B%5Cgamma%20_d%7D%7BG_s%20%5Cgamma%20_w%7D%20%3D%20%5Cfrac%7B20%7D%7B2.7%2A9.81%7D%20%3D%200.76%20%5C%20m%5E3%5C%5C%5C%5CVoid%20%5C%20volume%2C%20V_v%20%3D%201-0.76%20%3D%200.24%20%5C%20m%5E3)
![Void \ ratio, \ e = \frac{0.24}{0.76} = 0.32](https://tex.z-dn.net/?f=Void%20%5C%20ratio%2C%20%5C%20e%20%3D%20%5Cfrac%7B0.24%7D%7B0.76%7D%20%3D%200.32)
Submerged effective density is given as;
![\rho _b = \frac{\rho_w(\gamma_s -1)}{1+e}](https://tex.z-dn.net/?f=%5Crho%20_b%20%3D%20%5Cfrac%7B%5Crho_w%28%5Cgamma_s%20-1%29%7D%7B1%2Be%7D)
density of water (ρw) = 2.7 x 25 kN/m³ = 67.5 kN/m³, substitute this in the above equation;
![\rho _b = \frac{\rho_w(\gamma_s -1)}{1+e} = \frac{67.5(2.7 -1)}{1+0.32} = 86.93 \ kN/m^3](https://tex.z-dn.net/?f=%5Crho%20_b%20%3D%20%5Cfrac%7B%5Crho_w%28%5Cgamma_s%20-1%29%7D%7B1%2Be%7D%20%3D%20%20%5Cfrac%7B67.5%282.7%20-1%29%7D%7B1%2B0.32%7D%20%3D%2086.93%20%5C%20kN%2Fm%5E3)
Therefore, the submerged effective density is 86.93 kN/m³ or 8.693 Mg/m³