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Rashid [163]
3 years ago
8

A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust rever

sers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At lower speed, aerodynamic forces on the aircraft are small and may be neglected.
Physics
1 answer:
Amanda [17]3 years ago
4 0

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.

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Answer:

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Explanation:

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A brick sits on the top of a hill with a gravitational potential energy of 245 J. To determine the gravitational potential of th
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Answer:

The mass of the object, its acceleration due to gravity and the distance between the top of the hill and the ground level.

Explanation:

gravitational potential energy is the energy possessed by a body under influence of gravitational force by virtue of its position.

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3 years ago
(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of en
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Answer:

4.42 x 10⁷ W/m²

Explanation:

A = energy absorbed = 500 J

η = efficiency = 0.90

E = Total energy

Total energy is given as

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E = 500/0.90

E = 555.55 J

t = time = 4.00 s

Power of the beam is given as

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P = 138.88 Watt

d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m

Area of the circular spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

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Intensity of the beam is given as

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6 0
3 years ago
Read 2 more answers
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Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

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Answer:

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Explanation:

From the question we are told that

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     The coefficient of kinetic friction is  \mu_k  =  0.50

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        F_f  =  \mu_s  *  N

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        F_f  =   0.65 *  264.6

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Now for this  block to move the force require is  equal to F_f i.e

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=>    F  = 172 \ N

       

   

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