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:
<h2>The answer is 250 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume = 250 mL
density = 1 g/mL
So we have
mass = 250 × 1
We have the final answer as
<h3>250 g</h3>
Hope this helps you
Answer:
I=0.0361 kg.m^2
Explanation:
Torque is the rotational equivalent of a force
Torque= perpendicular distance r X Force F
Torque T = I(moment of inertia) X α (angular acceleration)
T= Iα
r= 0.0285m
F= 1.9 x 10^3
T=0.0285 x 1.9 x 10^3
T= 54.15Nm
I=T/α
I=54.15/150
I=0.361 kg.m^2
Answer:
The acceleration due to gravity is
times the value of g at the Earth’s surface.
(D) is correct option.
Explanation:
Given that,
Radius = 4000 miles
We need to calculate the gravitational force at surface
Gravitational force on the mass m on the surface of the earth
At r = R

....(I)
We need to calculate the gravitational force at height
Gravitational force on a mass m from the center of the earth,
At r = R + R = 2 R

....(II)
Dividing equation (II) by equation (I)


Hence, The acceleration due to gravity is
times the value of g at the Earth’s surface.
Answer:
270 m/s²
Explanation:
Given:
α = 150 rad/s²
ω = 12.0 rad/s
r = 1.30 m
Find:
a
The acceleration will have two components: a radial component and a tangential component.
The tangential component is:
at = αr
at = (150 rad/s²)(1.30 m)
at = 195 m/s²
The radial component is:
ar = v² / r
ar = ω² r
ar = (12.0 rad/s)² (1.30 m)
ar = 187.2 m/s²
So the magnitude of the total acceleration is:
a² = at² + ar²
a² = (195 m/s²)² + (187.2 m/s²)²
a = 270 m/s²