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3 years ago

ASAP

Physics

1 answer:

iris [78.8K]3 years ago

5 0

Answer:

A) Energy is dissipated into heat and sound energy due to Friction

B) The energy goes into heat and sound energy due to friction again, otherwise the cart would accelerate due to an unbalanced force. Therefore, we know there's friction, and the friction causes energy loss.

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What is matter? Give a grade 10/11/12 understanding.

lara [203]

Matter can be many things, but is mostly made of atoms. Atoms are small things that cannot be seen directly, as lights either passes through it or alters it. However, we know that atoms are made of 3 parts, those being the proton, neutron, and electron.

Protons have a positive electromagnetic charge.

Neutrons have no charge.

Electrons have a negative charge equal to the protons positive charge.

Protons and neutrons make up what's called the nucleus, which is orbited by the electrons.

Protons and neutrons also share another thing in common, that being their composition.

Until relatively recently, we thought that these were the smallest particles in the universe, and indestructible. However, modern discoveries have revealed that they are actually made of quarks and gluons.

These are actually indestructible, being part of the group that is elementary particles.

3 0

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:

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:

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