Study the distance-time graph, showing the distances that eight different things cover in 120 seconds or less. The letters below
correspond to the lines on the graph. For each object, calculate the average speed, and then match it with the closest correct answer.
A:
B:
C:
D:
E:
F:
G:
H:
A:
B:
C:
D:
E:
F:
G:
H:
1 answer:
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Answer:
26.9 Pa
Explanation:
We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:
(1)
where
is the cross-sectional area of the 1st section of the pipe
is the cross-sectional area of the 2nd section of the pipe
is the velocity of the 1st section of the pipe
is the velocity of the 2nd section of the pipe
In this problem we have:
is the velocity of blood in the 1st section
The diameter of the 2nd section is 74% of that of the 1st section, so
The cross-sectional area is proportional to the square of the diameter, so:
And solving eq.(1) for v2, we find the final velocity:
Now we can use Bernoulli's equation to find the pressure drop:
where
is the blood density
are the initial and final pressure
So the pressure drop is:
The motion of the ball on the vertical axis is an accelerated motion, with acceleration
The following relationship holds for an uniformly accelerated motion:
where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
So we can rewrite the equation as
from which we can isolate h
(1)
Now let's assume that
is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 
. So the maximum height of the second ball is
(2)
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
The following relationship holds for an uniformly accelerated motion:
where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
So we can rewrite the equation as
from which we can isolate h
Now let's assume that
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.