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3 years ago

When a football is kicked, the action and reaction forces do NOT cancel each other out because

Physics

1 answer:

AfilCa [17]3 years ago

6 0

Answer:

Explanation:

The kicker exerts more force of the football which is why the football moves when its kicked.

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Define malleability?<br>

ad-work [718]

Answer:

Explanation:

the quality or state of being malleable: such as. a : capability of being shaped or extended by hammering, forging, etc. the malleability of tin.

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3 years ago

Which type of hearing problem can be reduced which ordinary hearing aids

Paul [167]

Conduction deafness.

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3 years ago

How do you find the angle (to the nearest tenth) on your graph that produced the greatest range, based on the line of best fit?

VikaD [51]

Answer:

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3 years ago

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 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on

Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

$F_f = ma$ (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

$m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg$

And we can fidn the acceleration by using the formula:

$a=\frac{v-u}{t}$

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

$a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2$

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

$F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N$

Where the negative sign means the direction of the force is opposite to the motion of the box.

6 0

3 years ago