Answer:
What soil conditions favor the use of belled caissons?
Answer:
- where the bell can be unearthed from a solid surface.
- where the supporting stratum below the bottom of the caisson is impermeable to water movement.
What soil conditions favor piles over caissons?
Answer:
- non-cosheal soils
- subterranean water or excessive depth of bearing strata make caisson unworkable
What type of piles are especially well suited to repair or improvement of existing foundations ?
Answer:
Without hammering, minipiles or helical piles are placed which escapes much of the vibration and noise associated with traditional pile installation. for working close to existing buildings or for improving the exiting foundations where excessive vibration could damage exiting structures or noise may interfere with ongoing activities these piles are good options.
Why?
Their slenderness involves little or no displacement of the soil, thus minimizing the risk of disturbance to nearby foundations.
List and explain some cost thresholds frequently encountered in foundation design.
Answer:
building below the water table- site dewatering must occur, strengthening of slopes supper system must be done and waterproofing of the foundation all of which entails money
building near existing building - this requires underpinning(The process of reinforcing the base of an existing building or other structure underpins it.)
increase in column/wall load- building height determines the foundation depth
Answer:
M = 0.31 kg
Explanation:
This exercise must be done in parts, let's start by finding the speed of the set arrow plus apple, for this we define a system formed by the arrow and the apple, therefore the forces during the collision are internal and the moment is conserved
let's use m for the mass of the arrow with velocity v₁ = 20.4 m / s and M for the mass of the apple
initial instant. Just before the crash
p₀ = m v₁ + M 0
instant fianl. Right after the crash
p_f = (m + M) v
p₀ = p_f
m v₁ = (m + M) v
v = (1)
now we can work the arrow plus apple set when it leaves the child's head with horizontal speed and reaches the floor at x = 8 m. We can use kinematics to find the velocity of the set
x = v t
y = y₀ + t - ½ g t²
when it reaches the ground, its height is y = 0 and as it comes out horizontally,
0 = h - ½ g t²
t² = 2h / g
For the solution of the exercise, the height of the child must be known, suppose that h = 1 m
t =
t = 0.452 s
let's find the initial velocity
v = v / t
v = 8 / 0.452
v = 17.7 m / s
From equation 1
v = m / (m + M) v₁
m + M =
M = m + m \ \frac{v_1}{v}
we calculate
M = 0.144 + 0.144
M = 0.31 kg
Correct me if I’m wrong but I think it’s C
<span>6.95×10^28.
if this help then plz make it the brainiest
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