Download Physics Unit 5 Review

Physics
Unit 5: Fluids
1.
2.
3.
4.
5.
6.
7.
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13.
Meanings and concepts of terms like fluid, density, barometer, Pascal’s principle, Bernoulli’s principle, Archimedes’
principle, continuity equation, pressure, buoyant force, gauge pressure, absolute pressure, Poiseuille’s Law, laminar flow,
turbulent flow, osmosis, dialysis, diffusion
The density of mercury is 1.36 × 104 kg/m3. What is the mass of a 10-m3 sample of mercury?
The average density of the material in intergalactic space is approximately 2.5 × 10−27 kg/m3. What is the volume of a
gold sample, ρ = 19300 kg/m3, that has the same mass as 5 × 1024 m3 of intergalactic space?
A barometer is taken from the base to the top of a 10-m tower. Assuming the density of air is 1.29 kg/m3, what is the
measured change in pressure?
How much force does the atmosphere exert on one side of a vertical wall 10-m high and 20-m long?
A force of 500 N is applied to a hydraulic jack piston that is 0.01 m in diameter. If the piston which supports the load has
a diameter of 2 m, approximately how much mass can be lifted by the jack? Ignore any difference in height between the
pistons.
A balloon inflated with helium gas (density = 0.2 kg/m3) has a volume of 5 m3. If the density of air is 1.3 kg/m3, what is
the buoyant force exerted on the balloon?
Water enters a pipe of diameter 10 cm with a velocity of 5 m/s. The water encounters a constriction where its velocity is
20 m/s. What is the diameter of the constricted portion of the pipe?
A large tank is filled with water to a depth of 17 m. A spout located 8 m above
the bottom of the tank is then opened as shown in the drawing. With what
speed will water emerge from the spout?
A small crack occurs at the base of a 10 .0-m-high dam. The effective crack
17
area through which water leaves is 1.30 × 10−3 m2. Ignoring viscous losses,
what is the speed of the water flowing through the crack?
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Water flows through a pipe with radius 2 m and speed of 10 m/s. The density
−3
3
of water is 1000 kg/m and its viscosity is 1.002 × 10 Pa•s. Calculate the
Reynold’s number for this situation.
About how long will it take a perfume molecule to diffuse a distance of 2 m in
one direction in a room if the diffusion constant is 5 × 10−3 m2/s? Assume that the air is perfectly still.
The density of ice is 800 kg/m3; and the density of seawater is 900 kg/m3. A large iceberg floats in Arctic waters. What
fraction of the volume of the iceberg is exposed?
2.
3.
𝑘𝑔
𝜌 = 1.36 × 104 3 , 𝑉 = 10 𝑚3
𝑚
𝑚
𝜌=
𝑉
𝑘𝑔
𝑚
1.36 × 104 3 =
𝑚
10 𝑚3
𝟓
𝑚 = 𝟏. 𝟑𝟔 × 𝟏𝟎 𝒌𝒈
𝜌𝑠𝑝𝑎𝑐𝑒 = 2.5 × 10−27
𝑘𝑔
𝑚3
9.
, 𝜌𝑔𝑜𝑙𝑑 =
19300 3 , 𝑉𝑠𝑝𝑎𝑐𝑒 = 5 × 1024 𝑚3
𝑚
𝑚
𝜌=
𝑉
𝑘𝑔
𝑚
2.5 × 10−27 3 =
𝑚
5 × 1024 𝑚3
𝑚 = 0.0125 𝑘𝑔
𝑘𝑔 0.0125 𝑘𝑔
19300 3 =
𝑚
𝑉
𝑉 = 𝟔. 𝟒𝟖 × 𝟏𝟎−𝟕 𝒎𝟑
ℎ = 10 𝑚, 𝜌𝑎𝑖𝑟 = 1.29
2
𝑘𝑔
𝑚3
𝑃 = ℎ𝜌𝑔
𝑘𝑔
𝑚
) (9.8 2 ) = 𝟏𝟐𝟔 𝑷𝒂
𝑚3
𝑠
ℎ = 10 𝑚, ℓ = 20 𝑚
𝐹
𝑃=
𝐴
𝐹
1.01 × 105 𝑃𝑎 =
(10 𝑚)(20 𝑚)
𝐹 = 𝟐. 𝟎𝟐 × 𝟏𝟎𝟕 𝑵
𝐹1 = 500 𝑁, 𝑑1 = 0.01 𝑚, 𝑑2 = 2 𝑚
𝐹1 𝐹2
=
𝐴1 𝐴2
500 𝑁
𝐹2
=
2
𝜋(0.005 𝑚)
𝜋(1 𝑚)2
7
𝐹2 = 2.0 × 10 𝑁
𝑊 = 𝑚𝑔
𝑊 2.0 × 107 𝑁
𝑚=
=
= 𝟐. 𝟎𝟒 × 𝟏𝟎𝟔 𝒌𝒈
𝑔
9.8 𝑚/𝑠 2
𝑃 = (10 𝑚) (1.29
5.
6.
7.
8.
𝜌𝐻𝑒 = 0.2
𝑘𝑔
𝑚3
, 𝑉 = 5 𝑚3 , 𝜌𝑎𝑖𝑟 = 1.3
2
𝑘𝑔
𝑚
1 𝑎𝑡𝑚 + 0 + (1000 3 ) (9.8 2 ) (10 𝑚)
𝑚
𝑠
1
𝑘𝑔
= 1𝑎𝑡𝑚 + (1000 3 ) 𝑣22 + 0
2
𝑚
𝐽
𝑘𝑔 2
98000 2 = 500 3 𝑣2
𝑚
𝑚
𝑚
𝑣2 = 14
𝑠
𝑚
𝑘𝑔
11. 𝑟 = 2 𝑚, 𝑣 = 10 , 𝜌 = 1000 3 , 𝜂 = 1.002 ×
𝑠
𝑘𝑔
𝑚
) (10 ) (2 𝑚)
𝑠
𝑚3
𝑁𝑅 =
−3
1.002 × 10 𝑃𝑎 ⋅ 𝑠
𝑁𝑅 = 𝟑. 𝟗𝟗 × 𝟏𝟎𝟕
2 (1000
12. 𝑥 = 2 𝑚, 𝐷 = 5 × 10−3
2 𝑚 = √2 (5 × 10−3
4 𝑚2 = 10 × 10−3
𝐹𝐵 = 𝑚𝑎𝑖𝑟 𝑔
𝑚
𝜌=
𝑉
𝑘𝑔 𝑚𝑎𝑖𝑟
1.3 3 =
𝑚
5 𝑚3
𝑚𝑎𝑖𝑟 = 6.5 𝑘𝑔
𝑡 = 𝟒𝟎𝟎 𝒔
13. 𝜌𝑖𝑐𝑒 = 800
𝑠
𝑟2 = 0.025 𝑚
𝑑2 = 𝟎. 𝟎𝟓 𝒎
𝑠
𝑘𝑔
𝑚3
𝑚2
)𝑡
𝑠
𝑚2
⋅𝑡
𝑠
, 𝜌 = 900
𝑘𝑔
𝑚3
𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 =
𝑚
𝐹𝐵 = (6.5 𝑘𝑔) (9.8 2 ) = 𝟔𝟑. 𝟕 𝑵
𝑠
𝑚
𝑚
𝑑1 = 10 𝑐𝑚, 𝑣1 = 5 , 𝑣2 = 20
(𝜋(0.05 𝑚)2 ) (5
𝑚2
𝑥𝑟𝑚𝑠 = √2𝐷𝑡
𝑘𝑔
𝑚3
𝑚
10−3 𝑃𝑎 ⋅ 𝑠
2𝜌𝑣 𝑟
𝑁𝑅 =
𝜂
𝐹𝐵 = 𝑤𝑓𝑙
𝐴1 𝑣1 = 𝐴2 𝑣2
𝑘𝑔
𝑚3
1
1
𝑃1 + 𝜌𝑣12 + 𝜌𝑔ℎ1 = 𝑃2 + 𝜌𝑣22 + 𝜌𝑔ℎ2
2
2
𝑘𝑔
𝑚
1 𝑎𝑡𝑚 + 0 + (1000 3 ) (9.8 2 ) (17 𝑚)
𝑚
𝑠
1
𝑘𝑔
= 1 𝑎𝑡𝑚 + (1000 3 ) 𝑣22
2
𝑚
𝑘𝑔
𝑚
+ (1000 3 ) (9.8 2 ) (8 𝑚)
𝑚
𝑠
𝑁
𝑘𝑔 2
𝑁
166600 2 = (500 3 ) 𝑣2 + 78400 2
𝑚
𝑚
𝑚
𝑁
𝑘𝑔
88200 2 = (500 2 ) 𝑣22
𝑚
𝑚
𝒎
𝑣2 = 𝟏𝟑. 𝟑
𝒔
1
1
2
10. 𝑃1 + 𝜌𝑣1 + 𝜌𝑔ℎ1 = 𝑃2 + 𝜌𝑣22 + 𝜌𝑔ℎ2
𝑘𝑔
4.
ℎ1 = 17 𝑚, ℎ2 = 8 𝑚, 𝜌 = 1000
𝑠
𝑚
𝑚
) = (𝜋𝑟22 ) (20 )
𝑠
𝑠
𝜌𝑜𝑏𝑗
𝜌𝑓𝑙
𝑘𝑔
3
𝑚
𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 =
𝑘𝑔
900 3
𝑚
8
𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 = = 88.9 %
9
𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑒𝑥𝑝𝑜𝑠𝑒𝑑 = 1 − 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑
8
𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑒𝑥𝑝𝑜𝑠𝑒𝑑 = 1 −
9
𝟏
𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑒𝑥𝑝𝑜𝑠𝑒𝑑 = = 𝟏𝟏. 𝟏 %
𝟗
800