Answer:
a) t = 20 [s]
b) Can't land
Explanation:
To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = desacceleration = 5 [m/s^2]
t = time [s]
Note: the negative sign of the equation means that the aircraft slows down as it stops.
0 = 100 - 5*t
5*t = 100
t = 20 [s]
b)
Now we can find the distance using the following kinematics equation.
x - xo = distance [m]
x -xo = (0*20) + (0.5*5*20^2)
x - xo = 1000 [m]
1000 [m] = 1 [km]
And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land
The max is the largest it could get so ( ,0)
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
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D. How Much Energy Comes From A Bond Breaking.
