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murzikaleks [220]

3 years ago

14

Newton's third law states that for every action force there is an equal and opposite reaction force. An idiot in your class says

, "Wow that means everything cancels and nothing ever moves, it is all an illluussion! Wowwwwww" What statement best proves to him he is an idiot.
a) The equal and opposite forces act on different objects
b) If there is even a slight imbalance in the third law there will be a net force causing acceleration
C) he's right, all forces cancel, any motion I have ever seen is wrong

2 answers:

Otrada [13]3 years ago

4 0

A. Because the third laws say that for every action force the is an equal and opposite reaction force

Veronika [31]3 years ago

4 0

$\huge \mathfrak{Answer.... }$

The Correct Answer is :

B. if there is even a slight imbalance in third law there will be a net force causing acceleration.

A slight difference in the forces can result in acceleration of an object.

$\mathrm{✌TeeNForeveR✌}$

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Energy conservation

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Answer:

7.328m/s

Explanation:

Given parameters:

height of table = 0.68m

final velocity of the ball = 6m/s

Unknown:

Initial velocity of ball = ?

Solution:

To solve this problem, we are going to employ the appropriate motion equation.

We must understand that this fall occurs in the presence of gravity;

V = U + 2gH

Where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height of the pool table

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U = V - 2gH

U = 6 - (2 x 9.8 x 0.68) = 7.328m/s(deceleration)

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A car traveling at 100km/hr .how many hours will it take to cover a distance of 750 km

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3 years ago

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

B)

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

C)

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

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3 years ago

Hey what is you answering streak and what fact dose it say

Valentin [98]

Answer:

It’s 53:) Or - quit

Explanation:

That’s my strong streak

5 0

2 years ago

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