I am made of literally THOUSANDS of compounds ... too many to list here.
But the one compound that's most abundant, and actually comprises almost
80% of my entire beautiful body, is the compound DiHydrogen Oxide, with
the molecular formula H₂O . This compound is commonly known as "water".
No force is necessary to keep a moving object moving (in a straight line at a constant speed).
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.
The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds) = 4,000 foot-pounds of work.
If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds) = 645.2 foot-pounds per second.
The rate of doing work is called "power".
(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)
So back to our problem.
John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.
Well, 550 foot-pounds per second is called 1 "horsepower".
So as John runs up the steps to the balcony, he's doing the work
at the rate of
(645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)
= 1.173 Horsepower. GO JOHN !
(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________
Oh my gosh ! Look at #26 ! There are the metric units I was talking about.
Do you need #26 ?
I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.
a). 5
b). 750 Joules
c). 800 Joules
d). 93.75%
You're welcome.
And #27 is 0.667 m/s .
Answer:
The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]
Explanation:
We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.
For this case, we don't use the elevation difference and therefore those terms can be cancelled.
When the area has reduced the velocity of the fluid is increased but there is a drop pressure through the valve.