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3 years ago

PLEASE HELP!!! 30 POINTS

1 answer:

6 0

The impulse and the action and reaction law emit explain the sensation of the foot when kicking two different objects the answer in each case

a) kick the ball the sensation is not painful

b) kick the wall the feeling is painful

Newton's second law establishes a linear relationship between force and mass

F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration

Newton's third law or action and reaction law stable that the force acts on a pairs, the kicking of an object with a force F the object responds with a force of equal magnitude but opposite direction on us.

From Newton's second law we can derive a relationship between the impulse and the variation of the momentum

I = ∫ F dt = Δp

Where I is the impulse, t the time and Δp the variation of the momentum

The impulse is related to the sensation that the body receives due to the variation of the amount of movement, in this case we have two situations

a) Kick a soccer ball

The mass of the ball is small so the force applied by the foot creates a large acceleration without the foot having a significant decrease in foot speed. Consequently the impulse applied by the ball on the foot is small, the sensation is not painful

b) Kick a brick wall

In this case, the mass of the wall is high and when kicked it does not acquire any movement, consequently the reaction of the wall creates a high impulse on the foot that causes its speed to quickly decrease to zero, the physical sensation of this reduction in quantity movement is painful.

In conclusion, the impulse and the law of action and reaction allow us to find the answer in each case.

a) kick the ball feeling painless

b) kick the wall painful feeling

learn more about the action and reaction law and impulse here:

brainly.com/question/6677486

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Write one claim about this graph<br><br> Short answer

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The United States has a more significant economy than most countries around the world.

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3 years ago

Emma is working in a shoe test lab measuring the coefficient of friction for tennis shoes on a variety of surfaces. The shoes ar

Rashid [163]

Answer:

Explanation:

Force of friction = μ N , where μ is coefficient of friction , N is normal force on the body .

a )

Given,

Normal force N = 400 N

Force of friction = 300 N

μ = coefficient of static friction = ?

Putting the values ,

300 = 400 μ

μ = .75

b )

Normal force N = 400 N

Force of friction = 200 N

μ = coefficient of kinetic friction = ?

Putting the values ,

200 = 400 μ

μ = .50

c ) see attached file .

7 0

3 years ago

What is the net force (magnitude and direction) of the system below?

Crank

The net force (magnitude and direction) of the system given in the question is 40 N horizontal to the right

<h3>How to determine the net force</h3>

Case 1 (Net force between up and downward force)

Force up (Fu) = 50 N
Force down (Fd) = 30 N
Net force 1 (F1) = ?

F1 = Fu - Fd

F1 = 50 - 30

F1 = 20 N up

Case 2 (Net force between right and left)

Force right (Fr) = 60 N
Force left (Fl) = 20 N
Net force 2 (F2) = ?

F2 = Fr - Fl

F2 = 60 - 20

F2 = 40 N right

SUMMARY

Net force between up and down = 20 N up
Net force between right and left = 40 N right

From the above, the net force between right and left (i.e 40 N) is greater than the net force between up and down (i.e 20 N)

Thus, the net force of the system will be 40 N horizontal to the right

Learn more about force:

brainly.com/question/14361879

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2 years ago

What is the focal length of 2.30 D reading glasses found on the rack in a drugstore?

irakobra [83]

Answer:

Focal length, f = 0.43 meters

Explanation:

It is given that,

Power of the reading glasses, P = 2.3 D

We need to calculate the focal length of the reading glasses. The relationship between the power and the focal length inverse i.e.

$P=\dfrac{1}{f}$

$f=\dfrac{1}{P}$

$f=\dfrac{1}{2.3\ D}$

f = 0.43 m

So, the focal length of the reading glasses found on the rack in a drugstore is 0.43 meters.

8 0

4 years ago

In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

8 0

3 years ago