Answer:
The settlement that is expected is 1.043 meters.
Explanation:
Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil
The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

where
'H' is the initial depth of the layer
is the Compression index
is the inital void ratio
is the initial effective stress at the depth
is the change in the effective stress at the given depth
Applying the given values we get

Answer:
a) 
b) 
Explanation:
Previous concepts
The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution".
Part a
Let X the random variable of interest. We know on this case that 
And we know the probability denisty function for x given by:

In order to find the cdf we need to do the following integral:

Part b
Assuming that
, then the density function is given by:

And for this case we want this probability:

And evaluating the integral we got:

Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ =
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ =
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
Answer:
the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Explanation:
Given that;
depth 1 = 71 ft
depth 2 = 10 ft
pressure p = 17 psi = 2448 lb/ft²
depth h = 71 ft - 10 ft = 61 ft
we know that;
p = P_air + yh
where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )
so we substitute;
p = 2448 + ( 49.3 × 61 )
= 2448 + 3007.3
= 5455.3 lb/ft³
= 37.88 psi
Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi