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2 years ago

the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point

Physics

1 answer:

Andreyy892 years ago

5 0

Answer:

The pressure at this point is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

$P=1.81\ kW$

We need to calculate the pressure at this point

Using formula of pressure

$P=F\times v$

$P=A_{1}P_{1}\times v_{1}$

Here, $A_{1}v_{1}=Q_{1}$

Where, v = velocity

Put the value into the formula

$1.81 =P_{1}\times0.124\times\dfrac{1}{60}$

$P_{1}=\dfrac{1.81\times60}{0.124}$

$P_{1}=875.80\ kPa$

$P_{1}=0.875\ mPa$

Hence, The pressure at this point is 0.875 mPa

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A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and if $A_1=\frac{\pi D_{1}^{2}}{4}$ , then the expression would be: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ ,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
But, $D_2=2D_1$ and $L_2=2L_1$ , substituting in the expression for $Q_2$ : $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the last declaration of $Q_2$ , it could be noted that the expressión inside the parenthesis is actually $Q_1$ , then: $Q_2=\frac{2^2}{2}(25W)=50W$ .
<u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>

7 0

2 years ago

A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l

artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, $\lambda_1=8\times 10^{-6}\ C$

linear charge density of the other charge distribution, $\lambda_2=-6\times 10^{-6}\ C$

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.r.\epsilon_0}$

where:

$\lambda=$ linear charge density

r = radial distance from the center of wire

$\epsilon_0=$ permittivity of free space

Therefore,

$E_1=E_2$

$\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}$

$\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}$

$\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}$

$x=0.0629\ m$

∴at y=6.29 cm the charge of the two distribution will be equal.

9 0

2 years ago

Which actions most likely cause the domains in a ferromagnetic material to align?

Alexxx [7]

Answer:

A ferromagnetic material is a temporary magnet. The domains in a ferromagnetic material are randomly arranged. Under certain actions, the domains align in a particular direction and the material acts as a magnet. The actions that can cause alignment of domains in a ferromagnetic material are:

rubbing the material against a magnet would cause the alignment of domains in the same direction as of the magnet.
passing electricity around the material would generate magnetic field which would cause domains to align along the direction of the field.
placing the material near a strong magnet would cause the alignment of domains in the direction of the field generated by the strong magnet.

Other actions like heating the material, placing the material in a magnetic field of opposite polarity and hitting the material would lead to demagnetization of the magnetic material.

8 0

2 years ago

Read 2 more answers

When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t

antoniya [11.8K]

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

3 0

2 years ago

A rock with density 1900 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the st

wel

Answer:

the tension T2 when the rock is completely immersed is T2 = 29.05 N

Explanation:

from Newton's second law

F= m*a

where F= force , m= mass , a= acceleration

when the rock is suspended ,a=0 since it is at rest. Then

T1 - m*g = 0 , T1= tension when suspended in air , g= gravity

assuming constant density of the rock

m= ρ rock *V , where ρ rock = density of the rock , V= volume

thus

T1= m*g = ρ rock *g*V

V= T1/(ρ rock *g)

when the rock is submerged in oil , it receives an upward force that equals the weight of the volume of displaced oil (V displaced). Since it is completely submerged the volume displaced is the volume of the rock V=Vdisplaced

When the rock is at rest , then

F= m*a=0

T2 + ρ oil *g*V displaced - ρ rock *g*V =0

T2 = ρ rock *g*V - ρ oil *g*V = g*V (ρ rock - ρ oil)

T2 = g*V (ρ rock - ρ oil) = T1/(ρ rock *g) *g * (ρ rock - ρ oil)

T2 = T1 * (ρ rock - ρ oil)/ρ rock

replacing values

T2 = 48 N * (1900 kg/m3- 750 kg/m3)/ 1900 kg/m3 = 29.05 N

T2 = 29.05 N

3 0

2 years ago