Un autocar que circula a 81 km/h frena uniformemente con una aceleración de -4,5 m/s2.
1 answer:
Answer:
a)
b) imagen adjunta
Explanation:
a) Primero debemos hacer la conversión de 81 km/h a m/s, esto es 22.5 m/s.
Ahora, usando la ecuacion cinemática, en un movimiento acelerado tenemos:
Queremos encontrar la posición hasta detenerse, osea vf = 0.
b) Para este caso el gráfico se encuentra adjunto.
Espero que te sirva de ayuda!
You might be interested in
Answer:
= 85.89 ° C
Explanation:
The linear thermal expansion process is given by
ΔL = L α ΔT
For the three-dimensional case, the expression takes the form
ΔV = V β ΔT
Let's apply this equation to our case
ΔV / V = -0.507% = -0.507 10-2
ΔT = (ΔV / V) 1 /β
ΔT = -0.507 10⁻² 1 / 1.15 10⁻³
ΔT = -4.409
–T₀ = 4,409
= T₀ - 4,409
= 90.3-4409
= 85.89 ° C
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