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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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A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A
zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

V_{orbital} = \sqrt{\frac{GM_E}{R}}

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}

V_{orbital} = 5591.62m/s

V_{orbital} = 5.591*10^3m/s

PART B) The period of satellite is given as,

T = 2\pi \sqrt{\frac{r^3}{Gm_E}}

T = \frac{2\pi r}{V_{orbital}}

T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}

T = 238.61min

PART C) The gravitational force on the satellite is given by,

F = ma

F = \frac{1}{4} mg

F = \frac{270*9.8}{4}

F = 661.5N

5 0
3 years ago
Overcoming an object inertia always requires an
Zinaida [17]
Opposite force in the opposite.
3 0
3 years ago
What is the Gravitational Potential Energy of a 30 kg box lifted 1.5 meters off the ground?
juin [17]

Answer:

<h2>441 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 30 × 9.8 × 1.5

We have the final answer as

<h3>441 J</h3>

Hope this helps you

6 0
3 years ago
Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal accelerati
jek_recluse [69]

Answer:

a=4\ m/s^2

Explanation:

Given that,

Radius, r = 2 m

Velocity, v = 1 m/s

We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2

So, the magnitude of centripetal acceleration is 4\ m/s^2.

5 0
2 years ago
Why aren’t the Appalachian Mountains still as tall as the Himalayas?
stealth61 [152]

Answer:

mountains are limited in their theoretical height by several processes. First is isostasy: the bigger a mountain gets, the more it weighs down its tectonic plate, so it sinks lower. ... Bottom line: mountains can get taller than Mount Everest in earth gravity, like the Appalachians probably did—but not much taller.

3 0
2 years ago
Read 2 more answers
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