An airplane is flying at a constant speed in a positive direction. It slows down when it approaches the airport where it's going to land. this is an example of negative acceleration (D).
Answer:
t = 3/8 seconds
Explanation:
h=-16t^2 - 10t+6
h= 0 when it hits the ground
0=-16t^2 - 10t+6
factor out a -2
0= -2(8t^2 +5t -3)
divide by -2
0 = (8t^2 +5t -3)
factor
0=(8t-3) (t+1)
using the zero product property
8t-3 = 0 t+1 =0
8t = 3 t= -1
t = 3/8 t= -1
t cannot be negative ( no negative time)
t = 3/8 seconds
50*5=250
Momentum will be 250kgm/s^2
Answer:
Speed; v = 17 m/s
Explanation:
We are given;
Radius; r = 110m
Angle; θ = 15°
Now, we know that in circular motion,
v² = rg•tanθ
Thus,
v = √(rg•tanθ)
Where,
v is velocity
r is radius
g is acceleration due to gravity
θ is the angle
Thus,
v = √(rg•tanθ) = √(110 x 9.8•tan15)
v = √(288.85)
v = 17 m/s
