What is the value of work done on an object when a 0.1x102–newton force moves it 30 meters and the angle between the force and t
2 answers:
Answer:
A) 2.7 × 102 joules
Explanation:
The work done by a constant force is given by:
where
F is the magnitude of the force
d is the distance over which the force is applied
is the angle between the direction of the force and the displacement
In this problem, we have:
d=30 m
Substituting into the equation, we find the work done
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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:
560 kg m/s
Explanation:
First of all, we have to find the velocity of the runner, which is given by the ratio between the distance covered (400 m) and the time taken (50 s):
And now we can calculate the average momentum of the runner, which is equal to the product between the mass of the runner (70 kg) and its velocity, that we have previously calculated: