Download Exercise 12.1 Exercise 12.17

Workshop 2 (Chapter 12)
Exercise 12.1
On a part­time job, you are asked to bring a cylindrical iron rod of length 83.9 cm and diameter 2.60 cm from a
storage room to a machinist.
Part A
Calculate the weight of the rod.
ANSWER:
w
= 34.1 N Correct
Part B
Will you need a cart?
ANSWER:
yes
no
Correct
Exercise 12.17
There is a maximum depth at which a diver can breathe through a snorkel tube because as the depth increases, so
does the pressure difference, tending to collapse the
diver's lungs. Since the snorkel connects the air in the
lungs to the atmosphere at the surface, the pressure
inside the lungs is atmospheric pressure.
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Workshop 2 (Chapter 12)
Part A
What is the external­internal pressure difference when the diver's lungs are at a depth of 6.1 m (about 20 f t)?
Assume that the diver is in fresh water. (A scuba diver breathing from compressed air tanks can operate at
greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to
match the external pressure of the water.)
Express your answer using two significant figures.
ANSWER:
P
= 6.0×104 Pa Correct
Exercise 12.23
Part A
For the hydraulic lift shown in the figure , what must be the ratio of the diameter of the vessel at the car to the
diameter of the vessel where the force F1 is applied so that a 1510 kg car can be lifted with a force F1 of just
120 N ?
ANSWER:
Dcar
DF
= 11.1
1
Correct
W
± Buoyant Force Conceptual Question
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Workshop 2 (Chapter 12)
A rectangular wooden block of weight W floats with exactly one­half of its volume below the waterline.
Part A
What is the buoyant force acting on the block?
Hint 1. Archimedes' principle
The upward buoyant force on a floating (or submerged) object is equal to the weight of the liquid
displaced by the object.
Mathematically, the buoyant force Fbuoyant on a floating (or submerged) object is
F buoyant = ρgV
,
where ρ is the density of the fluid, V is the submerged volume of the object, and g is the acceleration
due to gravity.
Hint 2. What happens at equilibrium
The block is in equilibrium, so the net force acting on it is equal to zero.
ANSWER:
2W
W
1
2
W
The buoyant force cannot be determined.
Correct
Part B
The density of water is 1.00 g/cm3 . What is the density of the block?
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Workshop 2 (Chapter 12)
Hint 1. Applying Archimedes' principle
In Part A, you determined that the buoyant force, and hence the weight of the water displaced, is equal
to the weight of the block. Notice, however, that the volume of the water displaced is one­half of the
volume of the block.
Hint 2. Density
The density ρ of a material of mass m and volume V is
ρ=
m
V
.
ANSWER:
2.00 g/cm3
between 1.00 and 2.00 g/cm3
1.00 g/cm3
between 0.50 and 1.00 g/cm3
0.50 g/cm3
The density cannot be determined.
Correct
Part C
Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g
of mass. What is the mass of the wooden block?
Hint 1. How to approach the problem
When you add the extra mass on top of the block, the buoyant force must change. Relating the mass to
the new buoyant force will allow you to write an equation to solve for m.
Hint 2. Find the new buoyant force
With the 20­g mass on the block, twice as much of the block is underwater. Therefore, what happens to
the buoyant force on the block?
ANSWER:
The buoyant force doubles.
The buoyant force is halved.
The buoyant force doesn't change.
Hint 3. System mass
Before the 20­g mass is placed on the block, the mass of the system is just the mass of the block (m),
and this mass is supported by the buoyant force. With the 20­g mass on top, the total mass of the
system is now m + 20 g.
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Workshop 2 (Chapter 12)
Hint 4. The mass equation
The original buoyant force was equal to the weight W of the block. If the new buoyant force Fb is twice
the old buoyant force, then
F b = 2W = 2mg
,
where m is the mass of the block.
The new buoyant force must support the weight of the block and the mass, so according to Newton's
2nd law
.
By setting these two expressions for Fb equal to each other, you can solve for m.
F b = (m + 20)g
ANSWER:
40 g
20 g
10 g
Correct
Part D
The wooden block is removed from the water bath and placed in an unknown liquid. In this liquid, only one­third
of the wooden block is submerged. Is the unknown liquid more or less dense than water?
ANSWER:
more dense
less dense
Correct
Part E
What is the density of the unknown liquid ρ unknown ?
Express your answer numerically in grams per cubic centimeter.
Hint 1. Comparing densities
From Part C, the mass of the block is 20 g. In water, one­half of the block is submerged, so one­half of
the volume of the block times the density of the water must be equivalent to 20 g. In the unknown
substance, one­third of the block is submerged, so one­third of the volume of the block times the
density of the unknown substance must be equivalent to 20 g. You can use these observations to write
an equation for the density of the unknown substance.
Hint 2. Setting up the equation
The buoyant force on the block in water Fb,water is the same as the buoyant force in the unknown liquid F
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Workshop 2 (Chapter 12)
F b,unknown
,
F b,water = F b,unknown
,
so
ρ
water
g(
V
2
) = ρ unknown g(
V
3
),
where ρ water is the density of water, ρ unknown is the density of the unknown liquid, V is the volume of
the entire block, and g is the acceleration due to gravity.
ANSWER:
ρ unknown
= 1.5 g/cm3 Correct
Exercise 12.38
Water is flowing in a pipe with a varying cross­sectional area, and at all points the water completely fills the pipe. At
point 1 the cross­sectional area of the pipe is 0.070 m2 , and the magnitude of the fluid velocity is 3.50 m/s.
Part A
What is the fluid speed at point in the pipe where the cross­sectional area is 0.105 m2 ?
Express your answer using two significant figures.
ANSWER:
v
= 2.3 m/s Correct
Part B
What is the fluid speed at point in the pipe where the cross­sectional area is 0.047 m2 ?
Express your answer using two significant figures.
ANSWER:
v
= 5.2 m/s Correct
Part C
Calculate the volume of water discharged from the open end of the pipe in 1.00hour.
= 880 m3
V
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