The distance of an illuminated object from its light source is decreased to 1/3 while the luminous flux remains constant. Which
2 answers:
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A) 20 L/s
B) 2.55 m/s
C) 10.20 m/s
D) 400.8 kPa
Explanation:
a)
In this problem, we know that the volume of the tank:
is filled in a time of
We also know that the volume flow rate of water in the pipe is constant in every point of the pipe, so it can be calculated directly from the two data above.
The volume of the tank, in liter, is
While the time, in seconds, is
Therefore, the volume flow rate in Liters per second is:
b)
The volume flow rate of water through the pipe can be also written as
where
A is the cross-sectional area of the pipe
v is the speed of the water
In section B, we have:
r = 0.050 m is the radius of section B
so, the cross-sectional area of section B is:
The volume flow rate in SI units is
Therefore, the speed of the water in section B is:
c)
As in part B), we know that the volume flow rate must remain constant through the entire pipe.
So, the volume flow rate in section A of the pipe is still
The radius of the pipe in section A is
Therefore, the cross-sectional area in section A of the pipe is
So, since we have
we can find the speed of water in section A:
d)
Here we want to find the gauge pressure in section B.
We know that:
is the pressure in section A
is the altitude of section A
is the speed of water in section A
is the speed of water in section B
We can write Bernoulli's equation:
where
is the water density
is the pressure in section B
And solving for pB, we find:
Which is