Answer:
If a truss buckles or overturns, it is usually because of the failure of an adjacent truss or its bracing. A steel truss in a fire may buckle and overturn because of expansion or weakening from the heat. Most truss failures are the result of broken connections. Photo 1 shows a set of parallel-chord wood trusses supporting a plywood floor deck.
Explanation:
Answer:
9 cm
Explanation:
The liquid on the bend will be affected by two accelerations: gravity and centripetal force.
Gravity will be of 9.81 m/s^2 pointing down at all points.
The centripetal acceleration will be of
ac = v^2/r
Pointing to the center of the bend (perpendicular to gravity).
The velocity will depend on the radius
v = (1 m^2/s) / r
Replacing:
ac = (1/r)^2 / r
ac = (1 m^4/s^2) / r^3
If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be
a = (-1/r^3 * i - 9.81 * j)
The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.
The potential energy of the gravity field is:
pg = g * h
The potential energy of the centripetal force is:
pc = ac * r
Then the potential field is:
p = -1/r^2 * - 9.81*h
Points on the surface at r = 1 m and r = 3 m have the same potential.
-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2
-1 - 9.81*h1 = -1/9 - 9.81*h2
-1 + 1/9 = 9.81 * (h1 - h2)
h1 - h2 = (-8/9) / 9.81
h2 - h1 = 0.09 m
The outer part will be 9 cm higher than the inner part.
Answer:
Explanation:
There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.
We are given the following information:
y₁ = 3 m
y₂ = 3 m − 2.4 m = 0.6 m
t₂ − t₁ = 0.4 s
a = -9.8 m/s²
t₀ = 0 s
v₀ = 0 m/s
We need to find y₀.
Use a constant acceleration equation:
y = y₀ + v₀ t + ½ at²
Evaluated at point 1:
3 = y₀ + (0) t₁ + ½ (-9.8) t₁²
3 = y₀ − 4.9 t₁²
Evaluated at point 2:
0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²
0.6 = y₀ − 4.9 t₂²
Solve for y₀ in the first equation and substitute into the second:
y₀ = 3 + 4.9 t₁²
0.6 = (3 + 4.9 t₁²) − 4.9 t₂²
0 = 2.4 + 4.9 (t₁² − t₂²)
We know t₂ = t₁ + 0.4:
0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)
0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))
0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)
0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)
0 = 2.4 − 3.92 t₁ − 0.784
0 = 1.616 − 3.92 t₁
t₁ = 0.412
Now we can plug this into the original equation and find y₀:
3 = y₀ − 4.9 t₁²
3 = y₀ − 4.9 (0.412)²
3 = y₀ − 0.83
y₀ = 3.83
Rounded to two significant figures, the height of the tree is 3.8 meters.
Answer:
Explanation:
Given
Airline flying at 34,000 ft.
Cabin pressurized to an altitude 8,000 ft.
We know that at standard condition ,density of air
We know that pressure difference
ΔP=ρ g ΔZ
Here ΔZ=34,000-8,000 ft
ΔZ=26,000 ft
ΔP=0.074 x 32.2 x 26,000
So pressure difference will be .
