Answer:
part a : <em>The dry unit weight is 0.0616 </em>
<em />
part b : <em>The void ratio is 0.77</em>
part c : <em>Degree of Saturation is 0.43</em>
part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>
Explanation:
Part a
Dry Unit Weight
The dry unit weight is given as
Here
is the dry unit weight which is to be calculated- γ is the bulk unit weight given as
- w is the moisture content in percentage, given as 12%
Substituting values
<em>The dry unit weight is 0.0616 </em><em />
Part b
Void Ratio
The void ratio is given as
Here
- e is the void ratio which is to be calculated
- is the dry unit weight which is calculated in part a
is the water unit weight which is 62.4 
or 0.04
- G is the specific gravity which is given as 2.72
Substituting values
<em>The void ratio is 0.77</em>
Part c
Degree of Saturation
Degree of Saturation is given as
Here
- e is the void ratio which is calculated in part b
- G is the specific gravity which is given as 2.72
- w is the moisture content in percentage, given as 12% or 0.12 in fraction
Substituting values
<em>Degree of Saturation is 0.43</em>
Part d
Additional Water needed
For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as
Here
is the zero air unit weight which is to be calculated
- is the water unit weight which is 62.4 or 0.04
- G is the specific gravity which is given as 2.72
- w is the moisture content in percentage, given as 12% or 0.12 in fraction
Now as the volume is known, the the overall weight is given as
As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.
Answer: Well they could go down a hill to gain more kinetic energy, or the answer can just be B. He can pedal harder to increase the rate to 10 meters/second. I hope I helped you.
Explanation:
The electric field at a distance r from the charged particle is given by :
k is electrostatic constant
if r = 2 m, electric field is given by :
If r = 1 m, electric field is given by :
Dividing equation (1) and (2) we get :
So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.
Answer:
Since we have identical diodes we can use the equation:
And replacing we have:
Since we know that 1 mA is drawn away from the output then the real value for I would be
And for this case the value for 
would be:
And the output votage on this case would be:
And the net change in the output voltage would be:
Explanation:
For this case we have the figure attached illustrating the problem
We know that the equation for the current in a diode id given by:
![I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}](https://tex.z-dn.net/?f=%20I_D%20%3D%20I_s%20%5Be%5E%7B%5Cfrac%7BV_D%7D%7BV_T%7D%7D%20-1%5D%20%5Capprox%20I_S%20e%5E%7B%5Cfrac%7BV_D%7D%7BV_T%7D%7D)
For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode 
and each voltage is the same v for each diode, so then:
Since we have identical diodes we can use the equation:
And replacing we have:
Since we know that 1 mA is drawn away from the output then the real value for I would be
And for this case the value for would be:
And the output votage on this case would be:
And the net change in the output voltage would be:
Answer:
1) Measuring cylinder
2) Top Pan Balance
Explanation:
To find the density of the pebble Nabil should first accurately measure the mass of the pebble using the top pan balance
He should then fill the displacement can till the water reaches the outlet pipe at the to of the cylinder
He should then place the pebble into the cylinder carefully ans allow the water that would start coming out to stop
The volume of water in the displaced cylinder is then measured and the density is found by dividing the mass of the pebble by the volume of water in the measuring cylinder
