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Unformatted text preview: 3 3 ) = 6.20 ×10−2 m 624 FLUIDS 3 3 The weight of the shell is W = ρgg(V2 − V1) so R1 = R2 − (3/4)m/(π ρg), and R1 = 3 3m 1
1 =
−
4π ρ w ρg 3 3 (1.00 kg ) 1
1
− 3
3
4π
2.60 × 103 kg/m3 1.00 × 10 kg/m = 5.28 × 10−2 m 103. REASONING AND SOLUTION The force exerted by a pressure is F = PA and is
perpendicular to the surface. The force on the outside of each half of the roof is 133 N/m 2 4
F = (10.0 mm Hg) 1 mm Hg (14.5 m × 4.21 m ) = 8.11 × 10 N Adding the forces vectorially gives for the vertical force
Fv = 2F cos 30.0° = 1.41 × 105 N
and for the horizontal force Fh = 0 N. The net force is then 1.41×105 N .
The direction of the force is downward , since the horizontal components of the forces on
the halves cancel. 104. REASONING AND SOLUTION
a. Since the volume flow rate, Q = Av, is the same at each point, and since v is greater at the
lower point, the upper hole must have the larger area . b. Call the upper hole number 1 and the lower hole number 2 (the surface of the water is
position 0). Take the zero level of potential energy at the bottom hole and then write
Bernoulli's equation as
2 2 P1 + (1/2)ρv1 + ρgh = P2 + (1/2)ρv2 = P0 + ρg(2h)
in which P1 = P2 = P0 and from which we obtain v1 = 2 gh and v2 = 4 gh or v2 / v1 = 2 Chapter 11 Problems 625 Using the fact that Q1 = A1v1 = Q2 = A2v2 , we have v2/v1 = A1/A2. But since the ratio of the
2 2 areas is A1/A2 = r1 /r2 , we can write that
r1
r2 = A1
A2 = v2
v1 = 4 2 = 1.19 CHAPTER 12 TEMPERATURE AND HEAT ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (e) On each scale there are 100 degrees between the ice and steam points, so the size of the
degree is the same on each scale.
2. (b) According to Equation 12.2, the change in length ∆L of each rod is given by ∆L = αL0∆T, where α is the coefficient of linear expansion, L0 is the initial length, and ∆T is the change in
temperature. Since the initial length and the change in temperature are the same for each rod,
the rod with the larger coefficient of linear expansion has the greater increase in length as the
temperature rises. Thus, the aluminum rod lengthens more than the steel rod, so the rods will
meet to the right of the midpoint. 3. 1.2 × 10−3 cm
4. (d) In Arrangement I cooling allows the ball to pass through the hole. Therefore, the ball
must shrink more than the hole, and the coefficient of linear thermal expansion of metal A
must be greater than that of metal B. In Arrangement II heating allows the ball to pass
through the hole. Therefore, the coefficient of linear thermal expansion of metal C must be
greater than that of metal A.
5. (c) The gap expands as the temperature is increased, in a way similar to that of a hole when it
expands according to the coefficient of linear thermal expansion of the surrounding material.
In this case the surrounding material is copper.
6. (b) The change in volume ∆V is given by Equation 12.3 as ∆V = βV0∆T, where β is the
coefficient of volume thermal expansion, V0 is the initial volume, and ∆T is the change in
temperature. Since the sphere and the cube are made from the same material, the coefficient
of volume expansion is the same for each. Moreover, the temperature change is the same for
each. Therefore, the change in volume is proportional to the initial volume. The initial
volume of the cube is greater, since the sphere would fit within the cube. Thus, the change in
volume of the cube is greater.
7. (a) To keep the overflow to a minimum, the container should be made from a material that
has the greatest coefficient of volume thermal expansion and filled with a liquid that has the
smallest coefficient of volume thermal expansion. That way, when the full container is
heated, the cavity holding the liquid will expand more and the liquid will expand less, both
effects leading to a reduced amount of overflow.
8. 77 C° Chapter 12 Answers to Focus on Concepts Questions 627 9. (e) The heat Q required to raise the temperature of a mass m of material by an amount ∆T is
given by Equation 12.4 as Q = cm∆T, where c is the specific heat capacity of the material.
Since the material is the same in all cases, the specific heat capacity is the same. What
matters is the product of m and ∆T. Since this product is the same in all cases, the amount of
heat needed is also the same.
10. (c) The samples cool as heat is removed from each one. However, the temperature change
that results as heat is removed is different. The heat Q that must be removed to lower the
temperature of a mass m of material by an amount ∆T is given by Equation 12.4 as
Q = cm∆T, where c is the specific heat capacity of the material. Solving for ∆T gives
∆T = Q/(cm). For a given amount of heat removed, the fall in temperature is inversely
proportional to the product cm. The sample (sample A) with the largest value of cm will
experience the smallest drop in temperature. The sample (sample B) with the smallest value
of cm will experience the largest drop in temperature.
11. 49 °C
12. 21.6 °C
13. 6.75 kg
14. 0.38 kg
15. (d) Technique A is the way water normally freezes at normal atm...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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