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
Gravitational potential energy is the higher it us above the ground the more gravitational potential energy it holds.Sphere 2 says it has three times the mass of sphere 1. Therefore the answer is Sphere 2 since it was raised three time the mass of sphere 1. and the rest of the answer choices dont make sense.
At constant temperature and volume the pressure of a gas is directly proportional to the number of moles of the gas. PV =n RT, where R is the universal gas constant. A change in pressure depends with the number of moles of the gas, such that if the number of moles increases then there are many vibrations and collision of the gas molecules with the walls of a container thus increasing the pressure and vice versa.
Answer:
it would be 3
Explanation:
because you have to divide the length by the height of the incline.
