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Virty [35]
3 years ago
5

A glass pipe system has a very corrosive liquid flowing in it (think hydrofluoric acid, say). The liquid will destroy flow meter

s, but you need to know the flow rate. One way of measuring the flow rate is to add a fluorescent dye to the liquid at a known concentration, and then downstream activate the dye by UV light and then measure the dye concentration by emitted light. If the dye is added at 1.00 g/s, and the dye concentration downstream is 0.050% by mass, what is the unknown flow rate in kg/h
Physics
1 answer:
Soloha48 [4]3 years ago
5 0

A glass pipe system has a very corrosive liquid flowing in it (think hydrofluoric acid, say). The liquid will destroy flow meters, but you need to know the flow rate. One way of measuring the flow rate is to add a fluorescent dye to the liquid at a known concentration, and then downstream activate the dye by UV light and then measure the dye concentration by emitted light. If the dye is added at 1.00 g/s, and the dye concentration downstream is 0.050% by mass, what is the unknown flow rate in kg/h

 glass

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Pls help me w this, I've been doing this since 5 minutes ago​
natka813 [3]

Answer:

t=0.0625s

Explanation:

F=number of swings/time taken

DATA

Frequency=4.0Hz

number of swings from Q to R

=1/4

time taken=?

Frequency=number of swings/time taken

make t the subject of the formula

t=n/f

substitute the given date

t=0.25/4.0

t=0.0625s

option A is collect

4 0
3 years ago
As you stand near a railroad track, a train passes by at a speed of 31.7 m/s while sounding its horn at a frequency of 218 Hz. W
Darya [45]

Explanation:

Given that,

Frequency of train horn, f = 218 Hz

Speed of train, v_t = 31.7 m/s

The speed of sound, V = 344 m/s (say)

The speed of the observed person, V_o=0\ m/s

(a) When the train approaches you, the Doppler's effect gives the frequency as follows :

f'=f(\dfrac{V}{V-v_t})\\\\f'=218\times (\dfrac{344}{344-31.7})\\\\f'=240.12\ Hz

(b) When the train moves away from you, the Doppler's effect gives the frequency as follows :

f'=f(\dfrac{V}{V+v_t})\\\\f'=218\times (\dfrac{344}{344+31.7})\\\\f'=199.6\ Hz

Hence, this is the required solution.

6 0
4 years ago
The electric potential at a position located a distance of 20.7 mm from a positive point charge of 8.60×10-9C and 15.1 mm from a
max2010maxim [7]

Answer:

q2 = -4.35*10^-9C

Explanation:

In order to find the values of the second charge, you use the following formula:

V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}       (1)

V: electric potential = 1.14 kV = 1.14*10^3 kV

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1: charge 1 = 8.60*10^-9 C

q2: charge 2 = ?

r1: distance to the first charge = 20.7mm = 20.7*10^-3 m

r2: distance to the second charge = 15.1mm

You solve the equation (1) for q2, and replace the values of the other parameters:

q_2=\frac{r_2}{k}[V-k\frac{q_1}{r_1}]=\frac{Vr_2}{k}-\frac{q_1r_2}{r_1}\\\\q_2=\frac{(1.14*10^3V)(15.1*10^{-3}m)}{8.98*10^9Nm^2/C^2}-\frac{(8.60*10^{-9}C)(15.1*10^{-3}m)}{20.7*10^{-3}m}\\\\q_2=-4.35*10^{-9}C

The values of the second charge is -4.35*10^-9C

8 0
3 years ago
A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied
erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

\to \sigma y  =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\

Calculating Shear Transverse:

\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}

        = \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi

= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\  in

\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\

       = [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi

8 0
3 years ago
A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionle
Radda [10]

Answer:

a) x=0  %T=0,   b) x= A %T=100%,   c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

          x = A cos (wt + Ф)

where angular velocity is related to frequency and period

         w = 2π f = 2π / T

we can write the equation of the oscillation

         x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

         % T = 0

b) for x = A

 let's find the angle

      cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

            % T = 100%

c) the point of maximum compression x = -A

let's look for the angles

      cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

                % T = 50%

6 0
3 years ago
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