Answer:
1.44
Explanation:
0.1mm x 365 = 3.65mm
3.65mm into inches is 1.44
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.
Answer:
γ = 0.06563 N / m
9.78% difference
Explanation:
Solution:-
- Surface tension is the ability of any fluid to resist any external force which causes a decreases in surface area of the impact area due to inward compressive forces. These compressive forces occur due to cohesive nature of the fluid molecules.
- Mathematically, surface tension ( γ ) is defined as the force felt per unit length by the fluid.
γ = F / L
Where,
F: Force imparted
L: The length over which force is felt
- We are given the mass ( M ) of ( n = 100 ) water droplets to e 3.78 g. The mass of a single droplet ( m ) can be evaluated as follows:
m = M / n
m = 3.78 / 100
m = 0.0378 g
- The force ( F ) imparted by a single drop of water from the burette can be determined from the force balance on a single droplet. Assuming the distance over which the drop falls is negligible and resistive forces are negligible. Then the only force acting on the droplet is due to gravity:
F = m*g
F = 0.0378*9.81*10^-3
F = 0.000370818 N
- The length over which the force is felt can be magnified into a circular area with diameter equal to that of a single droplet ( d ). The circumferential length ( L ) of the droplet would be as follows:
L = π*d
L = π*( 0.0018 )
L = 0.00565 m
- Then the surface tension would be:
γ = F / L
γ = 0.000370818 / 0.00565
γ = 0.06563 N / m
- The tabulated value of water's surface tension is given as follows:
γa = 0.07275 N/m
- We will determine the percentage difference between the value evaluated and tabulated value as follows:

- The %difference between is within the allowable practical limits of 10%. Hence, the evaluated value ( γ = 0.06563 N / m ) can be accepted with 9.78% error.
I use the impulse momentum formula.
the 4.0 kilogram ball requires more force to stop