Help, It has multiple answers for this question.

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago

1.00kg of ice at -24 degrees Celsius is placed in contact with a 1.00kg block of a metal at 5.00 degrees Celsius. They come to e

Alona [7]

Answer:

C₂ = 2.22 KJ/kg °C

Explanation:

Since, both objects are in thermal contact. Therefore, the law of conservation of energy tells us that:

$Heat\ Lost\ by\ Metal\ Block = Heat\ Gained\ by\ Ice\\m_{1}C_{1} \Delta T_{1} = m_{2}C_{2} \Delta T_{2}$

where,

m₁ = mass of ice = 1 kg

C₁ = specific heat of ice = 2.04 KJ/kg.°C

ΔT₁ = Change in Temperature of Ice = -8.88°C - (-24°C) = 15.12°C

m₂ = mass of metal block = 1 kg

C₂ = specific heat of metal = ?

ΔT₂ = Change in Temperature of Metal Block = 5°C - (-8.88°C) = 13.88°C

Therefore, using these values in the equation, we get:

$(1\ kg)(2.04\ KJ/kg.^0C)(15.12\ ^0C) = (1\ kg)C_{2}(13.88\ ^0C) \\C_{2} = \frac{30.84\ KJ}{13.88\ kg/^0C}$

4 0

3 years ago

Read 2 more answers

I WILL MARK U BRAINLIEST IF U CAN ANSWER THIS QUESTION!!

sergey [27]

Answer:

The answer is (d) is totally reflected

Explanation:

Using the total internal reflection concept.

This concept state that:

"If the angle of incidence is greater than the critical angle then the light will not refract but it will reflect back totally"

Using the above statement, as incident angle is 42 degree and critical angle is 41 degree it means incident angle is greater than the critical angle. Therefore, light will not refract but it will totally reflected.

5 0

1 year ago

Can someone help me with this free body diagram problem?

zavuch27 [327]

Lol I just finished this

5 0

3 years ago

A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.90 m/s2 . At 20.0s after blastoff, t

Brilliant_brown [7]

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

$y=\frac{1}{2}at^2$

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

$y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m$

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

$t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s$

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}$

6 0

3 years ago