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3 years ago

The mass of an electron is 9.11 x 10-31 kg. If you want to increase the speed of the

1 answer:

5 0

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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

A sprinter runs 50m. You are given a stopwatch. Describe how you would measure: 1. the average speed of the sprinter over the en

Oksanka [162]

Answer:

Explanation:

Distance = 50 m

1. To find the average speed, first start the stop watch as the sprinter starts running and then stop it when he reaches the finish line.

Now note the time taken by the sprinter to run for 50 m.

The average speed of the sprinter is defined as the ratio of total distance covered to the total time taken.

Average speed = total distance / total time

2. To find the instantaneous speed, check the seed of the sprinter as he is at the finish line.

6 0

3 years ago

The gravitational pull between Pluto and the sun would be increased if

Troyanec [42]

Answer:

pluto was closer

Explanation:

6 0

3 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.23 2.23 times a second. A tack is stuck in the t

elena-14-01-66 [18.8K]

Explanation:

The given data is as follows.

Angular velocity ( $\omega$ ) = 2.23 rps

Distance from the center (R) = 0.379 m

First, we will convert revolutions per second into radian per second as follows.

= 2.23 revolutions per second

= $2.23 \times 2 \times 3.14 rad/s$

= 14.01 rad/s

Now, tangential speed will be calculated as follows.

Tangential speed, v = $R \times \omega$

= 0.379 x 14.01

= 5.31 m/s

Thus, we can conclude that the tack's tangential speed is 5.31 m/s.

8 0

3 years ago

The metric system is now known as the international system of units.<br> a. T<br> b. F

inn [45]

<span>The metric system is the oldest name for the international system of units. The answer is <u>a. True. </u>SI unit or the international systems of units are based on seven basic units; the meter, kilogram, second, ampere, Kelvin, candela and mole. All of these basic units are divided into multiples by a power of ten. For example in meters, 1 meter is equal to: 1000 millimeter, 100 centimeter, 10 decimeter, 0.1 decameter, 0.01 hectometer, 0.001 kilometer and so on and so forth.</span>

6 0

3 years ago