Ask question

Ask question

All categories

3 years ago

9

The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. Th

e water pressure in the line at the street is 130 psig. If I flow 10 gpm through the pipe, what pressure would I expect when I get to my house. My house is 10 ft higher in elevation than the water line at the street

1 answer:

jekas [21]3 years ago

8 0

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

= $130\times 6.894$

= $896.22 \ Kpa$

or,

= $896.22\times 10^3 \ Pa$

Z₂ = 10ft

= 3.05 m

$\delta$ = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒ $\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2$

On substituting the values, we get

⇒ $\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2$

⇒ $\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05$

⇒ $P_2=866330 \ P_a$

i.e.,

⇒ $=866330\times 0.000145$

⇒ $=126 \ Psi$

You might be interested in

Help<br><br><br> What is Gravitational Potenial Energy

frosja888 [35]

Gravitational potential energy is an object that possesses because of its position in a gravitational field

5 0

3 years ago

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$

$kx_1 = mg$

The extension of the spring due to the weight of the object on Moon is a value of $x_2$ , then

$kx_2 = mg_m$

Recall that gravity on the moon is a sixth of Earth's gravity.

$kx_2 = m\frac{g}{6}$

$kx_2 = \frac{1}{6} mg$

$kx_2 = \frac{1}{6} kx_1$

$x_2 = \frac{1}{6} x_1$

We have that the displacement at the earth was $x_1 = 0.3m$ , then

$x_2 = \frac{1}{6} 0.3$

$x_2 = 0.05m$

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0

3 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago

Suppose that you measure a galaxy's redshift, and from the redshift you determine that its recession velocity is 30,000 (3×10^4)

anyanavicka [17]

Answer:

1.4 billion light years away

Explanation:

v = Recessional velocity = 30000 km/s[/tex]

$H_0$ = Hubble constant = $\frac{65}{3.2\times 10^6}\ ly$

D = Distance to the galaxy

According to Hubble's law

$v=H_0D\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{3\times 10^{4}}{65}\times 3.2\times 10^6\\\Rightarrow D=1476923076.92307\ ly\\\Rightarrow D=1.4\times 10^9\ ly$

The galaxy is 1.4 billion light years away

5 0

3 years ago

A copper block rests 30.0 cm from the center of a steel turntable. The coefficient of static friction between the block and the

PIT_PIT [208]

Answer:

refer to the above attachment

3 0

2 years ago