Answer:
Gs = 2.647
e = 0.7986
Explanation:
We know that moist unit weight of soil is given as
![\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}](https://tex.z-dn.net/?f=%5Cgamma_m%20%5C%20or%5C%20bulk%5C%20density%20%3D%20%5Cfrac%7B%28Gs%2BSe%29%5Ctimes%20%5Cgamma_w%7D%7B%281%2Be%29%7D)
where,
= moist unit weight of the soil
Gs = specific gravity of the soil
S = degree of saturation
e = void ratio
= unit weight of water = 9.81 kN/m3
From data given we know that:
At 50% saturation,![\gamma_m = 16.62 kN/m3](https://tex.z-dn.net/?f=%20%5Cgamma_m%20%3D%2016.62%20kN%2Fm3)
puttng all value to get Gs value;
![16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}](https://tex.z-dn.net/?f=16.62%3D%20%5Cfrac%7B%28Gs%2B0.5%2Ae%29%5Ctimees%209.81%7D%7B%281%2Be%29%7D)
Gs - 1.194*e = 1.694 .........(1)
for saturaion 75%, unit weight = 17.71 KN/m3
![17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}](https://tex.z-dn.net/?f=17.71%20%3D%20%5Cfrac%7B%28Gs%2B0.75%2Ae%29%5Ctimes%209.81%7D%7B%281%2Be%29%7D)
Gs - 1.055*e = 1.805 .........(2)
solving both equations (1) and (2), we obtained;
Gs = 2.647
e = 0.7986
Answer:
well you could get some green goblin it disolves all the c rap in sink
Explanation:
Answer:
I hope following attachment will help you a lot!
Explanation:
Answer:
Explanation:
For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.
Answer: 24 pA
Explanation:
As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.
Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.
The resistance R of a given resistor, is expressed by the following formula:
R = ρ L / A
Replacing by the values for resistivity, L and A, we have
R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2
R = 2.1. 10¹¹ Ω
Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:
I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA