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
I think that your answer should be D) the state should make use of satellites and radars because they detect when a storm is coming
Answer:
a.
Explanation:
a
The property of air mass include
1) it must be large.
2) it must have relatively uniform properties.
3)it must travel as a recognizable entity.
It must have a warm front at its leading edge, is not necessarily a property of an air mass because Not all air masses have a warm front at their leading edge. There are five types of air masses and different types of the front can be developed.
Answer
given,
Capacitance of capacitor = 20.0-µF
Voltage = 150.0-V
inductance = 0.280 m H
a) the oscillation frequency of circuit
f = 2126.9 Hz
b) 
U = 0.225 J
c)Current in the inductor
instantaneous rate of change of current is equal to 535714 A/s
One minute has 60 seconds, One hour has 60 minutes and one day has 24 hours. Thus, 80 x 60 x 24 = 86,400 sec
if my answer helps you than mark me as brainliest
