Answer:
<h3>
2.3125m/s²</h3>
Explanation:
Using the equation of motion v² = u²+2aS
v is the final velocity = 120km/hr
120km/hr = 120 * 1000/1 * 3600 = 33.3m/s
u is the initial velocity = 0m/s
a is the acceleration
S is the distance covered = 240m
On substituting the given parameters
33.3² = 0²+2a(240)
33.3² = 480a
1110 = 480a
a = 1110/480
a = 2.3125m/s²
Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²
It was 40 I hope it helps
Given:
ρ = 1.18 kg/m³, density of air
v = 8 m/s, flow velocity
Q = 9 m³/s, volumetric flow rate
The minimum power required (at 100% efficiency) is
The actual power will be higher because 100% efficiency is not possible.
Answer: 339.8 W (nearest tenth)
In 1 hour, the hour hand sweeps across 1/12 of the clock's face. In 40 min, the hour hand travels (40 min)/(60 min) = 2/3 of the path it covers in an hour, so a total of 1/12 × 2/3 = 1/18 of the clock's face. This hand traces out a circle with radius 0.25 m, so in 40 min its tip traces out 1/18 of this circle's radius, or
1/18 × 2<em>π</em> (0.25 m) ≈ 0.087 m
The minute hand traverses (40 min)/(60 min) = 2/3 of the clock's face, so it traces out 2/3 of the circumference of a circle with radius 0.31 m:
2/3 × 2<em>π</em> (0.31 m) ≈ 1.3 m
The second hand completes 1 revolution each minute, so in 40 min it would fully trace the circumference of a circle with radius 0.34 m a total of 40 times, so it covers a distance of
40 × 2<em>π</em> (0.34 m) ≈ 85 m
