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
Answer:v=2.82 m/s
Explanation:
Given
initial velocity of ball(u)=3.6 m/s
ball rolls a distance(s) of 5 m
ball acceleration 
Velocity of ball when it is intercepted by Korean player




The answer is A. Hope this helps. :)
The kinetic Energy has to be the same as the Energy with which the spring was stratched:
1/2×m×v^2=1/2×D×s^2
Ekin=1/2×440×0.3^2= 1/2×440×0.09=19.8. ( Of course, the kinetic Energy is normally Not calculated Like this , but in this case both Formulas have the Same solution so you can Wright it Like This)
By the way, if the spring Had Not reached its Natural length yet, you wouldnt be right if you calculate it Like above, because a Part of the Energy is still in the spring : Espring=Ekin+Espring2