Answer:
Gs = 2.647
e = 0.7986
Explanation:
We know that moist unit weight of soil is given as
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,
puttng all value to get Gs value;
Gs - 1.194*e = 1.694 .........(1)
for saturaion 75%, unit weight = 17.71 KN/m3
Gs - 1.055*e = 1.805 .........(2)
solving both equations (1) and (2), we obtained;
Gs = 2.647
e = 0.7986
Answer:
a) 0.01
b) 150 cm^3/s
c) 300 cm^3/s
d) 25 cm^3/s
Explanation:
a) We know that :
Q=ΔP/R
R=8ηl/π*r^4
Givens:
r^2 = 0.1 r_1
Plugging known information to get :
Q=ΔP/R
=ΔP*π*r^4/8*η*l
Q_2/r_2^4 =Q_1/r_1^4
Q_2=Q_1/r_1^4*r_2^4
=Q_1/r_1^4*r*0.0001*r_1^4
Q_2 = 0.01
b) From the rate flow of the fluid we know that :
Q=ΔP/R (1)
F=η*Av/l (2)
R=8*ηl/π*r^4 (3)
<em>Where: </em>
ΔP is the change in the pressure .
r is the raduis of the tube .
l is the length of the tube .
η is the coefficient of the vescosity of the fluid .
R is the resistance of the fluid .
Givens: Q1 = 100 cm^3/s , ΔP= 1.5
Plugging known information into EQ.1 :
Q=ΔP/R
Q_2/ΔP2=Q_1/ΔP
Q_2=150 cm^3/s
c) we know that :
F = η*Av/l
can be written as :
ΔP = F/A = η*v/l
Givens: η_2 = 3η_1
Q=ΔP/R
Q=η*v/l*R
Q_2/η_2=Q_1/η_1
Q_2=300 cm^3/s
d) We know that :
Q=ΔP/R
R=8*ηl/π*r^4
Givens: l_2 = 4*l_1
Plugging known information to get :
Q=ΔP/R
Q=ΔP*π*r^4/8*ηl
Q_2/l_2=Q_1/l_1
Q_2 = 25 cm^3/s
Answer:
forward voltage triggering
temperature triggering
dv/dt triggering
light triggering
gate triggering
Then turning off;
Turn off is accomplished by a "negative voltage" pulse between the gate and cathode terminals
Explanation:
hope it helps
Answer should be C hopefully