(a) False. All points on the graph are positive (above the horizontal axis) so the blood always moves in the same direction. (You don't want blood to flow backwards in a vein or artery!) If this were a position graph, it would be a true statement because the slope changes from positive to negative at 0.25 s. The blood flow changes direction at about 0.25 seconds. (b) The acceleration of the blood is greatest at about 0.25 seconds. False. Acceleration at any point is the slope of the tanget line at that point. At 0.25 s, the slope is zero. True if it was an acceleration vs. time graph. (c) [3 pts.] The acceleration of the blood is greatest at about 0.10 seconds. True. Acceleration is determined from slope on a velocity graph, and the slope is greatest at about 01s

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i need an explanation on how the graph and how they came to the answers on a b and c and 4 a b

 

1.0
3. This is a typical graph of blood velocity versus
time during a single heartbeat. For each of the
following statements made about the graph, eval-
0.8
0.6
uate the statement as correct or incorrect. If the
0.4
statement is incorrect, identify the error (what
might the person have been thinking when mak-
ing the statement?) and correct it.
0.2
0.25
0.50
0.75
1.00
Time (s)
(a)
False. All points on the graph are positive (above the horizontal axis) so the blood always
moves in the same direction. (You don't want blood to flow backwards in a vein or artery!)
If this were a position graph, it would be a true statement because the slope changes from
positive to negative at 0.25 s.
The blood flow changes direction at about 0.25 seconds.
(b)
>The acceleration of the blood is greatest at about 0.25 seconds.
False. Acceleration at any point is the slope of the tanget line at that point. At 0.25 s, the
slope is zero. True if it was an acceleration vs. time graph.
(c) [3 pts.] The acceleration of the blood is greatest at about 0.10 seconds.
True. Acceleration is determined from slope on a velocity graph, and the slope is greatest
at about 0.1 s.
4. Consider a one-milligram blob of blood from the previous problem. Between 0.25s and 1.00s,
(a)
The graph is a straight line between 0.25 and 1.00 s, and the slope is 0.8/0.75 = 1.07, which
is our acceleration in m/s². Using v2 = v + 2AA¤, we have 0.8 m/s for v, and zero for
v, so solving for Ax gives us –0.3 m, so it travelled 30 cm (the negative is because we
calculated a displacement, but “how far" means we just want the magnitude).
how far does it travel?
(b)-
what is the force acting on it?
From F = ma, we have F = (1 mg)(1.07 m/s²) = (10–6 kg)(1.07 m/s²) =|1.07 µN
%3D
Blood velocity (m/s)
Transcribed Image Text:1.0 3. This is a typical graph of blood velocity versus time during a single heartbeat. For each of the following statements made about the graph, eval- 0.8 0.6 uate the statement as correct or incorrect. If the 0.4 statement is incorrect, identify the error (what might the person have been thinking when mak- ing the statement?) and correct it. 0.2 0.25 0.50 0.75 1.00 Time (s) (a) False. All points on the graph are positive (above the horizontal axis) so the blood always moves in the same direction. (You don't want blood to flow backwards in a vein or artery!) If this were a position graph, it would be a true statement because the slope changes from positive to negative at 0.25 s. The blood flow changes direction at about 0.25 seconds. (b) >The acceleration of the blood is greatest at about 0.25 seconds. False. Acceleration at any point is the slope of the tanget line at that point. At 0.25 s, the slope is zero. True if it was an acceleration vs. time graph. (c) [3 pts.] The acceleration of the blood is greatest at about 0.10 seconds. True. Acceleration is determined from slope on a velocity graph, and the slope is greatest at about 0.1 s. 4. Consider a one-milligram blob of blood from the previous problem. Between 0.25s and 1.00s, (a) The graph is a straight line between 0.25 and 1.00 s, and the slope is 0.8/0.75 = 1.07, which is our acceleration in m/s². Using v2 = v + 2AA¤, we have 0.8 m/s for v, and zero for v, so solving for Ax gives us –0.3 m, so it travelled 30 cm (the negative is because we calculated a displacement, but “how far" means we just want the magnitude). how far does it travel? (b)- what is the force acting on it? From F = ma, we have F = (1 mg)(1.07 m/s²) = (10–6 kg)(1.07 m/s²) =|1.07 µN %3D Blood velocity (m/s)
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