A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a
1 answer:
Answer:
24.3 degrees
Explanation:
A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:
Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.
Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.
So
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Explanation:
It is given that,
Mass of Millersburg Ferry, m = 13000 kg
Velocity, v = 11 m/s
Applied force, F = 10⁶ N
Time period, t = 20 seconds
(a) Impulse is given by the product of force and time taken i.e.
(b) Impulse is also given by the change in momentum i.e.
(c) For new velocity,
Hence, this is the required solution.
Answer:
98.15 lb
Explanation:
weight of plane (W) = 5,000 lb
velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s
wing area (A) = 200 ft^{2}
aspect ratio (AR) = 8.5
Oswald efficiency factor (E) = 0.93
density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}
Drag = 0.5 x ρ x x A x Cd
we need to get the drag coefficient (Cd) before we can solve for the drag
Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)
where
- induced drag coefficient (Cdi) =
(take note that π is shown as n and ρ is shown as
)
where lift coefficient (Cl)= =
= 0.245
therefore
induced drag coefficient (Cdi) = =
= 0.0024
- since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)
- Cd = 0.0024 + 0.0024 = 0.0048
Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.
Drag = 0.5 x ρ x x A x Cd
Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb
Answer:
y = 20.99 V / A
there is no friction y = 20.99 h
Explanation:
Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy
when the block of ice is submerged it is subjected to two forces its weight hydrostatic thrust
F_net= ∑F = B-W
the expression stop pushing is
B = ρ_water g V_ice
where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice
We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³
W = ρ-ice g V
F_net = (ρ_water - ρ_ ice) g V
this is the net force directed upwards, we can find the potential energy with the expression
F = -dU / dy
ΔU = - ∫ F dy
ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy
ΔU = - (ρ_water - ρ_ ice) g A y² / 2
we evaluate between the limits y = 0, U = 0, that is, the potential energy is zero at the surface
U_ice = (ρ_water - ρ_ ice) g A y² / 2
now we can use the conservation of mechanical energy
starting point. Ice depth point
Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2
final point. Highest point of the block
= U = m g y
as there is no friction, energy is conserved
Em₀ = Em_{f}
(ρ_water - ρ_ ice) g A y² / 2 = mg y
let's write the weight of the block as a function of its density
ρ_ice = m / V
m = ρ_ice V
we substitute
(ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y
y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A
let's substitute the values
y = 0.913 / (1 - 0.913) 2 V / A
y = 20.99 V / A
This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction
V / A = h
where h is the height of the block
y = 20.99 h