Answer:
runway use is 3307.8 feet
Explanation:
given data
velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s
time = 28 seconds
weight = 28000 lbs
to find out
How many feet of runway was used
solution
we will use here first equation of motion for find acceleration
v = u + at ..............1
here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time
put here value in equation 1
72.016 = 0 + a(28)
a = 2.572 m/s²
and
now apply third equation of motion
s = ut + 0.5×a×t² .......................2
here s is distance and u is initial speed and t is time and a is acceleration
put here all value in equation 2
s = 0 + 0.5×2.572×28²
s = 1008.24 m = 3307.8 ft
so runway use is 3307.8 feet
Answer:
21.5 m
Explanation:
A car has an initial speed of 31.4 km/hr
Convert to m/s
= 31.4 × 1000/3600
= 31,400/3600
= 8.722 m/s
Acceleration = 1.2 m/s^2
Time= 1.3 seconds.
Therefore the displacement can be calculated as follows
S= 8.722 × 1.3 + 1/2 × 1.2 × 1.3^2
= 11.34 + 1/2 × 20.28
= 11.34 + 10.14
= 21.5 m
Answer:
A larger force than 70 N will be required for the box to continue moving
Explanation:
A ramp is an inclined plane surface that is tilted to form a slope on its opposite sides
A ramp provides mechanical advantage or force amplification, by allowing less force to lift heavier load from having to move through a longer distance to reach a particular elevation when the slope of the ramp is gentle
Therefore, when the slope is steeper, and shorter, more force than 70 N will be required for the box to continue moving.
A = Δv/Δt
a = v2-v1/t2-t1
a = 60MS - 0MS / 6 seconds
a = 60 MS/ 6 seconds
a = 10 m/s^2
acceleration = change in velocity/change in time = 45-15 / 2
=30/2 = 15 m/s
