Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
<h2>
Angular acceleration is 80 rad/s²
</h2><h2>
Number of revolutions undergone is 1.02</h2>
Explanation:
We have equation of motion v = u + at
Initial angular velocity, u = 0 rad/s
Final angular velocity, v = 32 rad/s
Time, t = 0.40 s
Substituting
v = u + at
32 = 0 + a x 0.40
a = 80 rad/s²
Angular acceleration is 80 rad/s²
We have equation of motion s = ut + 0.5 at²
Initial angular velocity, u = 0 rad/s
Angular acceleration, a = 80 rad/s²
Time, t = 0.4 s
Substituting
s = ut + 0.5 at²
s = 0 x 0.4 + 0.5 x 80 x 0.4²
s = 6.4 rad
Angular displacement = 6.4 rad

Number of revolutions undergone is 1.02
Answer:
Therefore the resistance of the conductor is 175Ω
Explanation:
Resistance:
- Resistance of a metallic conductor is directly proportional to its length(l).
- Resistance of a metallic conductor is inversely proportional to its cross section area(A).
The notation sign of resistance is R.
The unit of resistance is ohm (Ω).
Therefore,

and



ρ is the proportional constant.
It is also known as resistivity of that metal.
Given ρ=35×10⁻⁶Ω-m
l= 20 m
A= 4.0×10⁻⁶m²

=175Ω
Therefore the resistance of the conductor is 175Ω
I believe it is green at 1,550