Physics 107 TUTORIAL ASSIGNMENT #6
Cutnell & Johnson, 7th edition
Chapter 12: Problem 17, 21, 37
Chapter 13: Problems 12, 31, 32, 34
Chapter 16: Problems 13, 27, 97
*17 The brass bar and the aluminum bar in the drawing are each attached to an immovable wall. At 28 °C the air gap between the rods is 1.3 x 10-3 m. At what temperature will the gap be closed?
*21 Consult Conceptual Example 5 for background pertinent to this problem. A lead sphere has a diameter that is 0.050% larger than the inner diameter of a steel ring when each has a temperature of 70.0 °C. Thus, the ring will not slip over the sphere. At what common temperature will the ring just slip over the ...view middle of the document...
The temperatures at the inside and outside surfaces are 27 °C and 0 °C, respectively. Each material has the same thickness and cross-sectional area. Find the temperature (a) at the plasterboard–brick interface and (b) at the brick–wood interface.
13 ssm The middle C string on a piano is under a tension of 944 N. The period and wavelength of a wave on this string are 3.82 ms and 1.26 m, respectively. Find the linear density of the string.
*27 ssm A transverse wave is traveling on a string. The displacement y of a particle from its equilibrium position is given by y = (0.021m)sin(25t – 2.0x). Note that the phase angle (25t - 2.0x) is in radians, t is in seconds, and x is in meters. The linear density of the string is
1.6 x 10-2 kg/m. What is the tension in the string?
*97 ssm The drawing shows a frictionless incline and pulley. The two blocks are connected by a wire (mass per unit length = 0.0250 kg/m) and remain stationary. A transverse wave on the wire has a speed of 75.0 m/s. Neglecting the weight of the wire relative to the tension in the wire, find the masses m1 and m2 of the blocks.
17. REASONING AND SOLUTION ΔL = αL0ΔT gives for the expansion of the aluminum
|ΔLA = αALAΔT |(1) |
and for the expansion of the brass
|ΔLB = αBLBΔT |(2) |
Taking the coefficients of thermal expansion for aluminum and brass from Table 10.1, adding Equations (1) and (2), and solving for ΔT give
The desired temperature is then
T = 28 °C + 21 C° = [pic]
21. REASONING AND SOLUTION The initial diameter of the sphere, ds, is
|ds = (5.0 × 10–4)dr + dr |(1) |
where dr is the initial diameter of the ring. Applying ΔL = αL0ΔT to the diameter of the
|Δds = αsdsΔT |(2) |
and to the ring gives
|Δdr = αrdrΔT |(3) |
If the sphere is just to fit inside the ring, we must have
ds + Δds = dr + Δdr
Using Equations (2) and (3) in this expression and solving for ΔT give
Substituting Equation (1) into this result and taking values for the coefficients of thermal expansion of steel and lead from Table 10.1 yield
The final temperature is
Tf = 70.0 °C − 29 C° = [pic]
37. [pic][pic] REASONING The cavity that contains the liquid in either Pyrex thermometer expands according to Equation 12.3, [pic]. On the other hand, the volume of...