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:
The speed of bullet and wooden bock coupled together, V = 22.22 m/s
Explanation:
Given that,
Mass of the bullet, m = 0.04 Kg
Mass of the wooden block, M = 0.5 Kg
The initial velocity of the bullet, u = 300 m/s
The initial velocity of the wooden block, U = 0 m/s
The final velocity of the bullet and wooden bock coupled together, V = 0 m/s
According to the conservation of linear momentum, the total momentum of the body after impact is equal to the total momentum before impact.
Therefore,
mV + MV = mu + MU
V(m+M) = mu
V = mu/(m+M)
Substituting the values in the above equation,
V = 0.04 Kg x 300 m/s / (0.04 Kg+ 0.5 Kg)
= 22.22 m/s
Hence, the speed of bullet and wooden bock coupled together, V = 22.22 m/s
Answer:
b = 0.6487 kg / s
Explanation:
In an oscillatory motion, friction is proportional to speed,
fr = - b v
where b is the coefficient of friction
when solving the equation the angular velocity has the form
w² = k / m - (b / 2m)²
In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction
let's call
w₀² = k / m
w² = w₀² - b² / 4m²
b² = (w₀² -w²) 4 m²
Let's find the angular velocities
w₀² = 5 / 0.1
w₀² = 50
w = 2π f
w = 2π 1
w = 6.2832 rad / s
we subtitute
b² = (50 - 6.2832²) 4 0.1²
b = √ 0.42086
b = 0.6487 kg / s
Answer:
a) 2.5 m/s. (In the opposite direction to the direction in which she threw the boot).
b) The centre of mass is still at the starting point for both bodies.
c) It'll take Sally 12 s to reach the shore which is 30 m from her starting point.
Explanation:
Linear momentum is conserved.
(mass of boot) × (velocity of boot) + (mass of sally) × (velocity of Sally) = 0
5×30 + 60 × v = 0
v = (-150/60) = -2.5 m/s. (Minus inicates that motion is in the opposite direction to the direction in which she threw the boot).
b) At time t = 10 s,
Sally has travelled 25 m and the boot has travelled 300 m.
Taking the starting point for both bodies as the origin, and Sally's direction as the positive direction.
Centre of mass = [(60)(25) + (5)(-300)]/(60+5)
= 0 m.
The centre of mass is still at the starting point for both bodies.
c) The shore is 30 m away.
Speed = (Distance)/(time)
Time = (Distance)/(speed) = (30/2.5)
Time = 12 s
Hope this Helps!!!
