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blsea [12.9K]
4 years ago
8

A bicycle traveled 150 meters west from point A to point B. Then it took the same route and came back to point A. It took a tota

l of 2 mins for the bicycle to return to point A. What is the average speed and average velocity of the bicycle?
A.
The average speed is 2.5 meters/second, and the average velocity is
2.5 meters/second east.

B.
The average speed is 0 meters/second, and the average velocity is
2.5 meters/second east.
C.
The average speed and average velocity are both 0 meters/second,
D.
The average speed is 2.5 meters/second, and the average velocity is
0 meters/second​
Physics
1 answer:
Katen [24]4 years ago
6 0
B i really
hope i helped
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An outside force

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Newton's law an object in motion stays in motion an object at rest stays at rest unless acted on by an outside force.

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If the velocity versus time graph of an object is a horizontal line, the object is
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Who invented the wedge simple machine and why?
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Explanation:

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What would happen to the amount of matter on earth if mass were not conserved during changes of state?
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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
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(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
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