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:
Please see attached graph.
Step-by-step explanation:
The equations for straight lines are given as:
i) y = x +4
ii) y= x-4
Yon can form a table for values of x and y that are true for an equation and use these values as coordinates ( x,y ) to plot the graphs and view the lines to select the correct labels for the equations.
For i)
y= x+4
x y coordinates
-3 1 (-3,1 )
-2 2 (-2,2)
-1 3 (-1,3)
0 4 ( 0,4)
1 5 ( 1,5)
2 6 ( 2,6)
3 7 ( 3,7)
Plot the points on a graph tool and draw the line. Do the same for the second equation to view both graphs as shown in the attached graph.
Answer:
770000
Step-by-step explanation:
im not sure
Answer:
760
Step-by-step explanation:
Answer:
D. The work shown above is correct, and
may not be simplified further.
Step-by-step explanation:
If you add all the exponents of y in first line then we get
4+4+4+4+4+3=23
Which is same as the exponent in original problem.
So first line is good.
Since it is 4th root so each
will turn into y when it moves out of the radical so final answer is correct.
Hence correct choice is "D. The work shown above is correct, and
may not be simplified further." is the correct answer.