Initial velocity of the plane is Vo = 0.
acceleration a = 1.3 m/s2
total distance = 2.5 km = 2500m
time taken to reach 2.5 km with 1.3m/s^2 acceleration = t
S = Vo t + 0.5 a t^2
2500 = 0 + (0.5*1.3* t^2)
t^2 = 3846.15
t = 62 s
the maximum velocity plan can reach within 62 s is Vt
Vt = Vo + a t
Vt = 0 + (1.3*62)
Vt = 80.6 m/s
Since 80.6 m/s is greater than 75 m/s, plane can use this runway to takeoff with required speed.
Answer:
The ball traveled 116.25 m when it hit the ground for the fifth term
Step-by-step explanation:
This is a geometric progression exercise and what we are asked to look for is the sum of a GP.
The ball was dropped from a height of 60 m. This means that the initial height of the ball is 60 m.
First value, a = 60
Each time it hit the ground, it bounced up 1/2 (half) of the height that it dropped.
This is the common ratio, r = 1/2 = 0.5
The number of terms it hits the ground is the number of terms in the GP.
number of terms, n = 5
The distance traveled by the ball when it hit the ground for the fifth term will be modeled by the equation:
Answer
B and E
Mark as brainly if possible!!
Answer:
35%
Step-by-step explanation:
Given data
Initial price= $80
Final price= $52
% decrease= initial-final/initial*100
% decrease= 80-52/80*100
% decrease= 28/80*100
% decrease=0.35*100
= 35%
Hence the decrease is 35%