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 higher you go the more potential energy there is, and the lower it is the more kinetic energy there is, so the more kinetic energy there is the higher the ball will bounce.
Answer:
AM has longer wavelength
Explanation:
The relation between the wavelength and teh frequency is given by
v = f x λ
Where, f is the frequency and λ be the wavelength.
It shows that the wavelength is inversely proportional to the frequency.
So, higher the frequency, smaller be the wavelength.
So, FM has high frequency than AM, thus, FM has lower wavelength as compared to AM.
The banking angle of the curved part of the speedway is determined as 32⁰.
<h3>
Banking angle of the curved road</h3>
The banking angle of the curved part of the speedway is calculated as follows;
V(max) = √(rg tanθ)
where;
- r is radius of the path
- g is acceleration due to gravity
V² = rg tanθ
tanθ = V²/rg
tanθ = (34²)/(190 x 9.8)
tanθ = 0.62
θ = arc tan(0.62)
θ = 31.8
θ ≈ 32⁰
Learn more about banking angle here: brainly.com/question/8169892
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