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:
Explanation:
Rocks tell us a great deal about the Earth's history. Igneous rocks tell of past volcanic episodes and can also be used to age-date certain periods in the past. Sedimentary rocks often record past depositional environments (e.g deep ocean, shallow shelf, fluvial) and usually contain the most fossils from past ages.
The star is FARTHER from Earth than the limit of our ability to measure parallax.
The NEAREST star outside the solar system has a parallax angle of 0.742 SECOND. That's like 0.000206 of a degree ! ALL other stars have SMALLER parallax.
I have no idea how they measure angles like these ... especially when the change in direction takes six months to happen !
B.) Action and reaction forces helps during lift off of a rocket.
Hope this helps!
Answer:
distance, x = 38.17 m
Given:
velocity of the football, v = 20.5 m/s
football projected at an angle, 
Period of flight, T = 2.15 s
Solution:
Since, the velocity is projected at some angle, so it can be resolved in horizontal and vertical components:
vertical component, 
horizontal component, 
the component responsible to cover certain distance is 
Therefore, the distance covered by the football, x is given by:
⇒ 
x = 38.17 m
