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
Answer:
The bottom of the sea is 25 m below sea level.
Explanation:
Given data
Mass = 6.1 ×
We know that Buoyant force on the tank is equal to gravity force of the tank.
1020 × = 6.1 ×
= 598039.21
We know that
= W × L × H
598039.21 = 300 × 80 × H
H = 25 m
Therefore the bottom of the sea is 25 m below sea level.
The expression of the electric flux is
Here,
Q = Total charge enclosed in the closed surface
= Permittivity due to free space
Rearranging to find the charge,
Replacing with our values we have finally
The charge enclosed by the box is 0.1684nC
The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.