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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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1 mile = 1609.344 meters 1 hour = 3600 seconds Initial speed of the getaway car = v1 = 0 m/s (At rest) Speed at which the getaway car explodes = v2 = 120 mph = 120 x (1609.344/3600) m/s = 53.…
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The speed of a motorcycle at a particular moment is it’s average speed or is instantaneous speed?
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4 years ago
What is the greatest eliminator of a species in terms of habitat destruction?
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3 years ago
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o
Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

U=-\frac{GMm}{r}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

and so, substituting:

R=6370 km\\h_A = 5970 km\\h_B = 21200 km

We find

\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

K=\frac{1}{2}\frac{GMm}{r}

So, the ratio between the two kinetic energies is

\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}

For satellite A, we have

E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J

For satellite B, we have

E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J

So, satellite B has the greater total energy (since the energy is negative).

(d) -2.57\cdot 10^8 J

The difference between the energy of the two satellites is:

E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J

4 0
3 years ago
2.
gizmo_the_mogwai [7]

Answer:

a) P1=100kpa

V1=6m³

V2=?

P2=50kpa

rearranging mathematically the expression for Boyle's law

V2=(P1V1)/P2=(100×6)/50=12m³

b) same apartment as in (a) but only the value of P2 changes

=> V2=(100×6)/40=15m³

Explanation:

since temperature is not changing we use Boyle's law. mathematically expressed as P1V1=P2V2

4 0
3 years ago
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