5B HW 3
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University of California, Los Angeles *
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Course
5B
Subject
Aerospace Engineering
Date
Jan 9, 2024
Type
Pages
14
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11/16/23, 9:13 PM
HW 3
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10862044
1/14
HW 3
Due: 11:59pm on Wednesday, November 8, 2023
To understand how points are awarded, read the
Grading Policy
for this assignment.
MCAT (R) Prep: Blood Pressure
At some instant, the blood pressure in the heart is 1.9
10
. Assume the density of blood is the same as water.
Part A
What is the pressure in an artery 0.20
below the heart, treating the blood as a nonviscous fluid?
ANSWER:
Correct
MCAT (R) Prep: Hydrostatic Weighing
Hydrostatic weighing, or "underwater weighing," is used to measure a patient's body composition. To do this measurement, the patient is weighed
while standing on a regular scale and then weighed while immersed in water. By comparing these weights, the density and body fat percentage of a
patient can be recorded. The weight of your patient on a regular scale is 650
and while immersed in water is 70
. The density of water is
1000
and assume that
= 10
.
Part A
What is the volume of your patient?
ANSWER:
Correct
Part B
If you were to unknowingly use salt water (
= 1025
) in place of fresh water, how would this affect your measurement of the patient's
density?
1.7
10
2.0
10
2.1
10
1.9
10
0.58
0.065
0.0070
0.058
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ANSWER:
Correct
Part C
What is the specific gravity of the patient?
ANSWER:
Correct
Part D
If the depth at which the underwater measurements were taken was doubled, what would be the patient's apparent weight?
ANSWER:
Correct
Part E
Suppose that while immersed your patient breathes through a thin tube. If the maximum pressure difference your patient's lungs can manage and
still breathe is 10,000
, what is the maximum depth to which your patient can be immersed?
ANSWER:
The patient's density would be measured lower than it actually is.
The patient's density would be measured correctly.
The patient's density would be measured higher than it actually is.
The patient's density would be measured as 0
.
0.89
0.11
1.12
1
140
35
17.5
70
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Correct
Force on a Goldfish Vector Drawing
A fishbowl contains a single goldfish and is filled with water to the level indicated.
Part A
At each of the designated points, rotate the given vector to indicate the direction of the force exerted by the water on either the inside of the
fishbowl (for points
A
and
B
) or the outside of the goldfish (for points
C
,
D
, and
E
).
The orientation of your vectors will be graded.
Hint 1.
Direction of the force due to the water
The water in the fishbowl will exert a force perpendicular to every surface it contacts. The pressure of the water, like the pressure in any
fluid, pushes the water outward, equally in all directions, exerting forces on whatever it comes into contact with.
ANSWER:
0.1
0.01
1
10
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Pressure with two liquids
shows a container with a cross section area of 15
in which a layer of water floats on top of a layer of mercury. A 1.0
wood block with a cross-
section area of 10
floats on the water. The water depth, measured from the bottom of the block, is 25
. A pressure gauge at the bottom of the
container reads 21
.
No elements selected
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Part A
What is the depth
of the mercury? Use
= 1000
as the density of water and
= 13,600
as the density of mercury.
Express your answer in centimeters.
Hint 1.
How to approach the problem
Hydrostatic pressure at depth
in a liquid of density
is
, where
is the pressure at the top of the liquid. This will need to
applied twice. Two other things are important. First, the pressure gauge reads
gauge pressure
, which is the pressure in excess of 1
.
Second, the weight of the block contributes to the pressure at the top of the water.
Hint 2.
Finding the gauge pressure at the top of the water
The pressure at the top of the water, at a point under the block, is 1
from the air plus weight per unit area of the block. But the gauge
pressure is the pressure in excess of 1
, so only the block contributes to the gauge pressure. What is the gauge pressure
at
the top of the water?
ANSWER:
Answer Requested
Hint 3.
Finding the gauge pressure at the bottom of the water
The gauge pressure at the top of the water, which you found in the previous hint, is the
term in the hydrostatic pressure equation for the
water. You know the depth and density of the water. What is the gauge pressure
at the bottom of the water?
Express your answer to four significant figures (to avoid round-off errors).
ANSWER:
Correct
Hint 4.
Finding the gauge pressure at the bottom of the mercury
The gauge pressure at the bottom of the water, which you found in the previous hint, is the
in the hydrostatic pressure equation for the
mercury. You know the density of the mercury. Write an expression for the gauge pressure
at the bottom of the mercury in terms of
= 9800
= 12250
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the unknown depth
.
Express your answer in terms of
, and express numerical values in SI units to four significant figures (to avoid round-off
errors).
ANSWER:
Correct
You know the numerical value of
so now you can solve for the depth.
ANSWER:
Correct
Problem 9.7
The deepest point in the ocean is 11
below sea level, deeper than Mt. Everest is tall.
Part A
What is the pressure in atmospheres at this depth?
Express your answer in atmospheres.
ANSWER:
Correct
Problem 9.6 - Enhanced - with Feedback
A 40-
-tall, 5.0-
-diameter cylindrical beaker is filled to its brim with water.
Part A
What is the total downward force on the bottom of the beaker?
Express your answer in newtons.
ANSWER:
Correct
=
= 6.6
= 1100
= 210
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Problem 9.13
Glycerin is poured into an open U-shaped tube until the height in both sides is 25
. Ethyl alcohol is then poured into one arm until the height of the
alcohol column is 25
. The two liquids do not mix.
Part A
What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is
1260
and the density of alcohol is 790
.
Express your answer with the appropriate units.
ANSWER:
Answer Requested
Problem 9.12 - Enhanced - with Feedback
An ocean-going research submarine has a 10.0-
-diameter window 8.20
thick. The manufacturer says the window can withstand forces up to
1.20×10
6
. The pressure inside the submarine is maintained at 1.0
.
Part A
What is the submarine's maximum safe depth?
Express your answer with the appropriate units.
ANSWER:
Correct
Here we learn how to calculate the maximum safe depth of a submarine if we know the size of its window and the force it can withstand.
Problem 9.16
A
block floats in water with its long axis vertical. The length of the block above water is 3.0
.
Part A
What is the block's mass density?
Express your answer with the appropriate units.
ANSWER:
= 9.3
= 1.5×10
4
= 670
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Correct
Problem 9.19
A sphere completely submerged in water is tethered to the bottom with a string. The tension in the string is one-third the weight of the sphere.
Part A
What is the density of the sphere?
Express your answer with the appropriate units.
ANSWER:
Correct
Problem 9.18 - Enhanced - with Hints and Feedback
Astronauts visiting a new planet find a lake filled with an unknown liquid. They have with them a plastic cube, 8.0
on each side, with a density of
840
. First they weigh the cube with a spring scale, measuring a weight of 21
. Then they float the cube in the lake and find that two-thirds of
the cube is submerged.
Part A
At what depth in the lake will the pressure be twice the atmospheric pressure of 85
?
Express your answer with the appropriate units.
Hint 1.
How to approach the problem
Apply the hydrostatic pressure equation to determine the depth at which the pressure is twice the atmospheric pressure. From the
information about the submerged part of the cube, one can determine the ratio between the densities of the cube and the liquid in the lake.
Also, note that the free-fall acceleration on the planet may not be the same as on the earth, but you can calculate it using Newton's second
law.
ANSWER:
Answer Requested
Problem 9.24 - Enhanced - with Video Solution
Styrofoam has a density of
.
= 750
= 140
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HW 3
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For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of
Floating bowling ball
.
Part A
What is the maximum mass that can hang without sinking from a 90.0-
-diameter Styrofoam sphere in water? Assume the volume of the mass
is negligible compared to that of the sphere.
Express your answer with the appropriate units.
ANSWER:
Answer Requested
Here we learn how to use Archimedes' principle to calculate the mass that can be supported by the buoyancy force acting on a styrofoam
sphere.
Submerged Sphere in a Beaker
A cylindrical beaker of height 0.100
and negligible weight is filled to the brim with a fluid of density
= 890
. When the beaker is placed on
a scale, its weight is measured to be 1.00
.
A ball of density
= 5000
and volume
= 60.0
is then submerged in the
fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a
stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration
due to gravity is
= 9.81
.
Part A
What is the weight
of the ball?
Express your answer numerically in newtons.
Hint 1.
Find the mass of the ball
Find the mass
of the ball.
Express your answer numerically in kilograms.
ANSWER:
Hint 2.
Converting between cubic meters and cubic centimeters
Since
, then
and
.
ANSWER:
= 267
= 0.300
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Part B
What is the reading
of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the
scale.
Calculate your answer from the quantities given in the problem and express it numerically in newtons.
Hint 1.
How to approach the problem
One way to answer this question is to compute the pressure at the bottom of the beaker. Since the walls of the beaker are vertical, the total
vertical force on the beaker due to the fluid is a result of this pressure. Alternatively, you can compute the weight of the water in the beaker
(recall that some water is lost over the edge when the ball is submerged) and add the force exerted by the ball on the water (which is
transmitted through the water to the beaker).
Hint 2.
Find the pressure at the bottom of the beaker
What is the pressure
of the fluid at the bottom of the beaker?
Express your answer numerically in pascals
ANSWER:
Hint 3.
Find the pressure before the ball is submerged
What was the pressure
of the fluid at the bottom of the beaker before the ball was submerged?
Express your answer numerically in pascals.
ANSWER:
Hint 4.
The force exerted by the ball on the water
According to Newton's 3rd law, the force exerted by the ball on the water is equal and opposite to the buoyant force exerted by the water
on the ball.
ANSWER:
Correct
The "extra force" that the ball exerts on the water is equal to the force that the water exerts on the ball, that is, the weight of the
displaced water. Therefore, the reading does not change.
Part C
What is the force
applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer
should be negative).
Express your answer numerically in newtons.
Hint 1.
How to approach the problem
= 2.94
= 873
= 873
= 1.00
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HW 3
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The ball is stationary. Therefore, the total force acting on it must be zero. The total force includes the force of the rod, the gravitational
force, and the buoyant force.
Hint 2.
Find the buoyant force
Using Archimedes' principle, find
, the buoyant force acting on the submerged ball.
Express your answer numerically in newtons.
ANSWER:
Hint 3.
Find the expression for the net force
Taking the positive
y
direction to be vertically upward, choose an equation correctly expressing the net force acting on the ball in terms of
the existing forces. Let
, the force exerted by the rod, be directed upward.
Hint 1.
How to approach this problem
There are three forces acting on the ball: the weight acting downward, the buoyant force acting upward, and the force from the rod,
assumed to be acting upward.
ANSWER:
ANSWER:
Correct
The force
does
act upward as one would expect because the ball is denser than the fluid and would have sunk if it weren't for the rod.
The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod,
and the beaker with fluid and submerged ball is placed on the scale.
Part D
What weight
does the scale now show?
= 0.524
= 2.42
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Express your answer numerically in newtons.
Hint 1.
How to approach the problem
It makes no difference how the ball and fluid are arranged inside the beaker. If there are no external forces, the weight reading would be
equal to the total weight of the beaker and its contents.
Hint 2.
Find the weight of the fluid in the beaker
What is the weight
of the fluid that overflowed the beaker when the ball was originally submerged?
Express your answer numerically in newtons.
ANSWER:
ANSWER:
Correct
Note that in this case it is not necessary to know the buoyant force acting on the ball in to answer the question.
MCAT (R) Prep: Coronary Heart Disease
Coronary heart disease is caused by a buildup of plaque in the arteries that supply blood to the heart. When the flow of blood to the heart is restricted
or blocked, the heart can be damaged due to a lack of oxygen. Further, if a large piece of plaque buildup gets dislodged from an artery wall, it can get
stuck in other arteries throughout the body, including arteries in the brain.
For simplicity, let's assume that the artery is cylindrical with a radius of
. The flow rate of a viscous fluid is given by Poiseuille's Law
.
Part A
If plaque buildup reduces the radius of the artery by a factor of 2, by what factor does the flow rate change?
ANSWER:
Correct
Part B
If the plaque buildup (modeled as a cylinder as well) completely blocks the artery, a pressure difference of
between each side of the clot builds
up. What is the force on the clot?
ANSWER:
= 0.524
= 3.42
The flow rate is 16 times the original.
The rate is 4 times the original.
The flow rate is now 1/16th of the original.
The flow rate is 1/4th of the original.
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Part C
If the clot breaks free and only experiences a force due to the blood pressure, what is the relationship between the pressure on each side of the
clot as it moves with a constant velocity toward the heart?
ANSWER:
Correct
Reading Quiz 9.10
Part A
Blood flows more slowly in the capillaries than in the aorta. Why?
ANSWER:
Correct
Even though an individual capillary is much narrower than the aorta, there are many of them, so their total area is larger than the aorta
by a large factor.
Video Tutor: Bernoulli's Principle-Venturi Tubes
There is a pressure differential in order to keep the clot moving.
The pressure on the side furthest from the heart is higher because the clot is moving toward the heart.
The pressure on the side closest to the heart is higher because the clot is moving toward the heart.
The pressure on each side is the same.
A capillary is narrower than the aorta.
Less blood flows through the capillaries per second than through the aorta.
The blood pressure in the capillaries is higher than in the aorta.
The total area of the capillaries is larger than the area of the aorta.
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HW 3
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First,
launch the video
below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video
window and answer the question on the right. You can watch the video again at any point.
Part A
A Ping-Pong ball hangs from a string that is attached to the rear view mirror of a parked car (see ). The driver side window is open, and the
passenger side window is closed. A strong breeze blows along
the driver side of the car. What happens to the Ping-Pong ball?
Assume the breeze is laminar and that Bernoulli’s equation
applies.
Hint 1.
While the breeze is blowing, the air speed inside the open driver side window will be higher than the air speed inside the closed passenger
side window. Think back to the video demonstration. What is the relationship between the speed of a fluid and the pressure it exerts?
ANSWER:
Correct
Since the air outside the driver side window is moving, the pressure on the driver side will be lower than the pressure on the passenger
side, where the air is initially still. The Ping-Pong ball will be pushed towards the driver side window (i.e. towards the area of lower
pressure).
Note that we have assumed that the flow is smooth. In a real car traveling at highway speeds, the motion of the air may be more
turbulent and hard to predict.
Score Summary:
Your score on this assignment is 73.5%.
You received 11.77 out of a possible total of 16 points.
The ball swings towards the passenger side.
The ball swings towards the driver side.
The ball does not move in either direction.