436_PartUniversity Physics Solution

436_PartUniversity Physics Solution - 13-28 Chapter 13...

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13-28 Chapter 13 13.101. IDENTIFY: Eq.(13.39) says 2/ TI m g d π = . SET UP: Let the two distances from the center of mass be 12 and . dd There are then two relations of the form of Eq. (13.39); with 22 1c m 1 2c m 2 and II m d d =+ . EXECUTE: These relations may be rewritten as ( ) 2 m 1 4 mgd T I md and ( ) 2 m 2 4. mgd T I md Subtracting the expressions gives () ( ) 2 2 2 1 2 1212 44 . mg d d T m d d m d d d d ππ −= = −+ Dividing by the common factor of ( ) md d and letting dd L + = gives the desired result. EVALUATE: The procedure works in practice only if both pivot locations give rise to SHM for small oscillations. 13.102. IDENTIFY: Apply m F =a # # to the mass, with 2 rad aa r ω == . SET UP: The spring, when stretched, provides an inward force. EXECUTE: Using 2 l for the magnitude of the inward radial acceleration, 2 0 mlk l l = , or 0 2 kl l km = . (b) The spring will tend to become unboundedly long. EXECUTE: As resonance is approached and l becomes very large, both the spring force and the radial acceleration become large. 13.103. IDENTIFY: For a small displacement x , the force constant k is defined by x Fk x = − . SET UP: Let 00 , so that and rR x 2 [] . bx bx FA e e −− =− EXECUTE: When x is small compared to 1 , b expanding the exponential function gives [(1 2 ) (1 )] , F A bx bx Abx ≈−− = corresponding to a force constant of 579 N/m Ab = . EVALUATE: Our result is very close to the value given in Exercise 13.40.
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14-1 F LUID M ECHANICS 14.1. IDENTIFY: Use Eq.(14.1) to calculate the mass and then use wm g = to calculate the weight. SET UP: / mV ρ = so = From Table 14.1, 33 7.8 10 kg/m . EXECUTE: For a cylinder of length L and radius R , 22 4 3 ( ) (0.01425 m) (0.858 m) 5.474 10 m . VR L ππ == = × Then 4 3 (7.8 10 kg/m )(5.474 10 m ) 4.27 kg, == × × = and 2 (4.27 kg)(9.80 m/s ) 41.8 N g = (about 9.4 lbs). A cart is not needed. EVALUATE: The rod is less than 1m long and less than 3 cm in diameter, so a weight of around 10 lbs seems reasonable. 14.2. IDENTIFY: Convert gallons to kg. The mass m of a volume V of gasoline is = . SET UP: 1 gal 3.788 L 3.788 10 m × . 3 1 m of gasoline has a mass of 737 kg. EXECUTE: 3 1 gal 1 m 45.0 mi/gal (45.0 mi/gal) 16.1 mi/kg 737 kg ⎛⎞ ⎜⎟ × ⎝⎠ EVALUATE: 1 gallon of gasoline has a mass of 2.79 kg. The car goes fewer miles on 1 kg than on 1 gal, since 1 kg of gasoline is less gasoline than 1 gal of gasoline. 14.3. IDENTIFY: / = SET UP: The density of gold is 19.3 10 kg/m × . EXECUTE: 333 6 3 (5.0 10 m)(15.0 10 m)(30.0 10 m) 2.25 10 m V −−− × × = × . 63 0.0158 kg 7.02 10 kg/m m V = × × . The metal is not pure gold. EVALUATE: The average density is only 36% that of gold, so at most 36% of the mass is gold. 14.4. IDENTIFY: Find the mass of gold that has a value of 6 $1.00 10 × . Then use the density of gold to find the volume of this mass of gold. SET UP: For gold, . The volume V of a cube is related to the length L of one side by 3 VL = . EXECUTE: 3 6 1 troy ounce 31.1035 10 kg ($1.00 10 ) 72.9 kg $426.60 1 troy ounce m × = .
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436_PartUniversity Physics Solution - 13-28 Chapter 13...

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