# Physics

posted by on .

I have been given this question and with very little knowledge of physics and a book and flex studies that I am not finding very explanitory. I am not sure I am on the correct path as the vlaues I have seem very large. I would appreciate any help/assistance anyone could give.
Question
You measure the blood pressure at the bicep and find that it is 140 mmHg systolic and 80mmHg diastolic. What would you expect the blood pressure to be at the top of the head given that this point is 45cm above the measurement point? Remember thisis a gauge pressure and give your answers in both mmHg and kPa. (The density of blood is about 1060 kgm^-3)

This is what I have done but I am so unsure on what I am doing.

1. Converted the systolic and diastolic blood pressure into Pa.
Systolic Blood pressure= 140mmHg x 133.322Pa/1mmHg= 18665 Pa
Diastolic Blood pressure= 80mmHg x 133.322Pa/1mmHg= 10665.76Pa

2. Due to being gauge pressure I then converted to actual pressure. (this is something I am very unsure if it is needed)
P1 actual systolic pressure= 18665Pa + 101325Pa = 119990Pa
P1 actual diastolic pressure= 10665.76Pa + 101325Pa = 111990.76Pa

3. Due to the change in hieght I named the top of the head P2.
P2 - P1=pg h2-h1
P2=P1+pgh
P2 systolic = 119990Pa + (1060kgm^-3 x 9.81 m s^-2 x 45m) = 587927Pa
P2 diastolic = 111990.76 + ( 1060kgm^-3 x 9.81 m s^-2 x 45m) = 579928Pa

4. I then converted the P2 into kPa.
P2 systolic kPa = 587927Pa x .1kPa/100Pa = 587.9kPa
P2 diastolic KPa = 579928Pa x .1kPa/100Pa = 579.9kPa

5. I then converted the P2 into mmHg. (this seems very large)
P2 systolic mmHg = 587927Pa x 1mmHg/133.322Pa = 4410mmHg
P2 diastolic mmHg = 579928Pa x 1mmHg/133.322Pa=4350mmHg

I have spent a lot of time trying to work this out. If you can please assist me I would be very greatful

• Physics - ,

Both pressures will be higher by pressure = (rho)*g*H when measured at elevation H higher than the arm. rho is the density of blood. Assume it is the same as water

Your numbers are much too high.

Atmospheric pressure is 760 mm Hg

450 mm higher elevation is equivalent to 45/13.6 = 33 mm Hg higher pressure, measured in mm Hg. The 13.6 factor id the ratio of the densities of mercury and water.

• Physics (correction) - ,

Pressures in the head will be lower, not higher, because of the higher elevation than the measurement point.

The number is also wrong:

It should be 45 mmHg/13.6 = 3.3 mmHg lower

I just wasn't thinking and apologize for the errors.

• Physics - ,

Thank you for your help. I am still unsure on the formulas I am meant to be using and when I am meant to be converting from mmHg to Pa.
The text passage contains a physics problem related to blood pressure and its variation with height, along with an attempt to solve it and some feedback on the solution.
single_question_with_answer

1. the passage contains a single mathematical question about calculating the expected blood pressure at the top of the head given the blood pressure at the bicep and the height difference.
2. an answer to the question is provided in the form of feedback, correcting the student's calculations and providing the correct approach and values.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Physics

posted by on .

I have been given this question and with very little knowledge of physics and a book and flex studies that I am not finding very explanitory. I am not sure I am on the correct path as the vlaues I have seem very large. I would appreciate any help/assistance anyone could give.
Question
You measure the blood pressure at the bicep and find that it is 140 mmHg systolic and 80mmHg diastolic. What would you expect the blood pressure to be at the top of the head given that this point is 45cm above the measurement point? Remember thisis a gauge pressure and give your answers in both mmHg and kPa. (The density of blood is about 1060 kgm^-3)

This is what I have done but I am so unsure on what I am doing.

1. Converted the systolic and diastolic blood pressure into Pa.
Systolic Blood pressure= 140mmHg x 133.322Pa/1mmHg= 18665 Pa
Diastolic Blood pressure= 80mmHg x 133.322Pa/1mmHg= 10665.76Pa

2. Due to being gauge pressure I then converted to actual pressure. (this is something I am very unsure if it is needed)
P1 actual systolic pressure= 18665Pa + 101325Pa = 119990Pa
P1 actual diastolic pressure= 10665.76Pa + 101325Pa = 111990.76Pa

3. Due to the change in hieght I named the top of the head P2.
P2 - P1=pg h2-h1
P2=P1+pgh
P2 systolic = 119990Pa + (1060kgm^-3 x 9.81 m s^-2 x 45m) = 587927Pa
P2 diastolic = 111990.76 + ( 1060kgm^-3 x 9.81 m s^-2 x 45m) = 579928Pa

4. I then converted the P2 into kPa.
P2 systolic kPa = 587927Pa x .1kPa/100Pa = 587.9kPa
P2 diastolic KPa = 579928Pa x .1kPa/100Pa = 579.9kPa

5. I then converted the P2 into mmHg. (this seems very large)
P2 systolic mmHg = 587927Pa x 1mmHg/133.322Pa = 4410mmHg
P2 diastolic mmHg = 579928Pa x 1mmHg/133.322Pa=4350mmHg

I have spent a lot of time trying to work this out. If you can please assist me I would be very greatful

• Physics - ,

Both pressures will be higher by pressure = (rho)*g*H when measured at elevation H higher than the arm. rho is the density of blood. Assume it is the same as water

Your numbers are much too high.

Atmospheric pressure is 760 mm Hg

450 mm higher elevation is equivalent to 45/13.6 = 33 mm Hg higher pressure, measured in mm Hg. The 13.6 factor id the ratio of the densities of mercury and water.

• Physics (correction) - ,

Pressures in the head will be lower, not higher, because of the higher elevation than the measurement point.

The number is also wrong:

It should be 45 mmHg/13.6 = 3.3 mmHg lower

I just wasn't thinking and apologize for the errors.

• Physics - ,

Thank you for your help. I am still unsure on the formulas I am meant to be using and when I am meant to be converting from mmHg to Pa.
user
The text passage contains a physics problem related to blood pressure and its variation with height, along with an attempt to solve it and some feedback on the solution.

TYPE: SINGLE_QUESTION_WITH_ANSWER

1. The passage contains a single mathematical question about calculating the expected blood pressure at the top of the head given the blood pressure at the bicep and the height difference.
2. An answer to the question is provided in the form of feedback, correcting the student's calculations and providing the correct approach and values.