Is a computer an open or closed system

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

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

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3 years ago

The diagram shows the electric field near conducting sphere a which of the following statements is correct

Ludmilka [50]

I really don’t know but I hope this helps

4 0

3 years ago

When the starter motor on a car is engaged, there is a 300 A cunent in the wires between the bauery and the motor. Suppose the w

docker41 [41]

Answer:

diameter of transmission is 3.473 mm

Explanation:

given data

current = 300 A

total wire length = 1.0 m

voltage drop = 0.55 V

solution

as here Resistivity of copper is = 1.68 × $10^{-8}$ ohm- m

so we get here resistance of wire that is

resistance = $\frac{V}{I}$ ..............1

R = $\frac{0.55}{310}$

R = 1.774 × $10^{-3}$ ohm

and now we get diameter of wire that is

R = $\frac{\rho l}{A}$ .............2

$\pi \frac{d^2}{4} = \frac{\rho l}{R}$

d² = $\frac{4\rho l}{\pi R}$ ...............3

put here value

d² = $\frac{4\times 1.68\times 10^{-8} \times 1}{\pi \times 1.774\times 10^{-3}}$

d = $\sqrt{12.064\times 10^{-6}}$

d = 3.473 × $10^{-3}$ m

d = 3.473 mm

so diameter of transmission is 3.473 mm

4 0

4 years ago

Occasionally, people can survive falling large distances if the surface they land on is soft enough. During a traverse of Eiger'

uranmaximum [27]

Answer:

4611.58 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.174 ft/s²

Equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32.174\times 516+0^2}\\\Rightarrow v=182.218\ ft/s$

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-182.218^2}{2\times 3.6}\\\Rightarrow a=-4611.58\ ft/s^2$

Magnitude of acceleration while stopping is 4611.58 ft/s²

8 0

3 years ago

Need answers quick !! will give brainliest !!

stiv31 [10]

Answer:

use the bowl of water because Earth's magnetic field is relatively weak. Allowing it to float freely on the water, allows the magnetized needle to freely react to Earth's magnetic field, causing it to align North to South. If you watched closely, the same end of the needle should always point to the North

Explanation:

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3 years ago