Soluciones fisica
Páginas: 8 (1891 palabras)
Publicado: 24 de noviembre de 2010
SOLUCIONES TAREA UNO
14.3: [pic] You were cheated.
14.7: [pic]
[pic]
14.9: a) [pic]
b) [pic]
14.15: With just the mercury, the gauge pressure at the bottom of the cylinder is[pic] With the water to a depth [pic], the gauge pressure at the bottom of the cylinder is[pic] If this is to be double the first value, then [pic]
[pic]
The volumeof water is [pic]
14.17: The force is the difference between the upward force of the water and the downward forces of the air and the weight. The difference between the pressure inside and out is the gauge pressure, so
[pic]
14.25: a) [pic]
b) [pic]
c) [pic]
The density of the block is [pic] Note that is the same as the average density of the fluiddisplaced, [pic]
14.26: a) Neglecting the density of the air,
[pic]
[pic]to two figures.
b)[pic]
14.51: a) [pic]
b) The gauge pressure at a depth of 15.0 [pic]below the top of the mercury column must be that found in part (a); [pic] which is solved for [pic]
14.64: The buoyant force on the mass A, divided by[pic](see Example 14.6), so the mass block is [pic]a)The mass of the liquid displaced by the block is [pic]so the density of the liquid is [pic]b) Scale D will read the mass of the block, [pic]as found above. Scale E will read the sum of the masses of the beaker and liquid, [pic]
SOLUCIONES TAREA DOS
14.29: [pic]
[pic]
[pic]
14.33: The hole is given as being “small,”and this may be taken to mean that the velocity of theseawater at the top of the tank is zero, and Eq. (14.18) gives
[pic]
=[pic]
[pic]
Note that y = 0 and [pic] were used at the bottom of the tank, so that p was the given gauge pressure at the top of the tank.
14.35: The assumption may be taken to mean that [pic]in Eq. (14.17). At the maximum height,[pic] and using gauge pressure for [pic](the water is open to theatmosphere), [pic]
14.39: The water is discharged at a rate of [pic] The pipe is given as horizonatal, so the speed at the constriction is [pic] keeping an extra figure, so the cross-section are at the constriction is [pic] and the radius is [pic]
14.41: Let point 1 be where [pic] and point 2 be where[pic] The volume flow rate has the value [pic] at all points in the pipe.
[pic][pic]
[pic]
[pic]
SOLUCIONES TAREA TRES
13.1: a) [pic][pic]
b) [pic][pic]
13.4: (a) From the graph of its motion, the object completes one full cycle in 2.0 s; its period is thus 2.0 s and its frequency [pic] (b) The displacement varies from [pic] so the amplitude is 0.20 m. (c) 2.0 s (see part a)
13.5: This displacement is [pic] of a period.
[pic]
13.7: a) [pic]13.15: The equation describing the motion is [pic]this is best found from either inspection or from Eq. (13.14) (Eq. (13.18) involves an infinite argument of the arctangent). Even so, [pic]is determined only up to the sign, but that does not affect the result of this exercise. The distance from the equilibrium position is [pic]
13.17: [pic][pic]
[pic]
13.19: [pic]
[pic][pic]13.23: a) Setting [pic] in Eq. (13.21) and solving for x gives [pic] Eliminating x in favor of v with the same relation gives [pic] b) This happens four times each cycle, corresponding the four possible combinations of + and – in the results of part (a). The time between the occurrences is one-fourth of a period or [pic]
13.27: a) [pic].
b) [pic]
c) [pic]
13.29: [pic][pic]
13.31: a) [pic]
b) [pic]
13.41: [pic] so for a different acceleration due to gravity [pic]
[pic]
13.43: Besides approximating the pendulum motion as SHM, assume that the angle
is sufficiently small that the length of the spring does not change while swinging in the
arc. Denote the angular frequency of the vertical motion as [pic]
[pic] which is solved for [pic]. But L is...
14.3: [pic] You were cheated.
14.7: [pic]
[pic]
14.9: a) [pic]
b) [pic]
14.15: With just the mercury, the gauge pressure at the bottom of the cylinder is[pic] With the water to a depth [pic], the gauge pressure at the bottom of the cylinder is[pic] If this is to be double the first value, then [pic]
[pic]
The volumeof water is [pic]
14.17: The force is the difference between the upward force of the water and the downward forces of the air and the weight. The difference between the pressure inside and out is the gauge pressure, so
[pic]
14.25: a) [pic]
b) [pic]
c) [pic]
The density of the block is [pic] Note that is the same as the average density of the fluiddisplaced, [pic]
14.26: a) Neglecting the density of the air,
[pic]
[pic]to two figures.
b)[pic]
14.51: a) [pic]
b) The gauge pressure at a depth of 15.0 [pic]below the top of the mercury column must be that found in part (a); [pic] which is solved for [pic]
14.64: The buoyant force on the mass A, divided by[pic](see Example 14.6), so the mass block is [pic]a)The mass of the liquid displaced by the block is [pic]so the density of the liquid is [pic]b) Scale D will read the mass of the block, [pic]as found above. Scale E will read the sum of the masses of the beaker and liquid, [pic]
SOLUCIONES TAREA DOS
14.29: [pic]
[pic]
[pic]
14.33: The hole is given as being “small,”and this may be taken to mean that the velocity of theseawater at the top of the tank is zero, and Eq. (14.18) gives
[pic]
=[pic]
[pic]
Note that y = 0 and [pic] were used at the bottom of the tank, so that p was the given gauge pressure at the top of the tank.
14.35: The assumption may be taken to mean that [pic]in Eq. (14.17). At the maximum height,[pic] and using gauge pressure for [pic](the water is open to theatmosphere), [pic]
14.39: The water is discharged at a rate of [pic] The pipe is given as horizonatal, so the speed at the constriction is [pic] keeping an extra figure, so the cross-section are at the constriction is [pic] and the radius is [pic]
14.41: Let point 1 be where [pic] and point 2 be where[pic] The volume flow rate has the value [pic] at all points in the pipe.
[pic][pic]
[pic]
[pic]
SOLUCIONES TAREA TRES
13.1: a) [pic][pic]
b) [pic][pic]
13.4: (a) From the graph of its motion, the object completes one full cycle in 2.0 s; its period is thus 2.0 s and its frequency [pic] (b) The displacement varies from [pic] so the amplitude is 0.20 m. (c) 2.0 s (see part a)
13.5: This displacement is [pic] of a period.
[pic]
13.7: a) [pic]13.15: The equation describing the motion is [pic]this is best found from either inspection or from Eq. (13.14) (Eq. (13.18) involves an infinite argument of the arctangent). Even so, [pic]is determined only up to the sign, but that does not affect the result of this exercise. The distance from the equilibrium position is [pic]
13.17: [pic][pic]
[pic]
13.19: [pic]
[pic][pic]13.23: a) Setting [pic] in Eq. (13.21) and solving for x gives [pic] Eliminating x in favor of v with the same relation gives [pic] b) This happens four times each cycle, corresponding the four possible combinations of + and – in the results of part (a). The time between the occurrences is one-fourth of a period or [pic]
13.27: a) [pic].
b) [pic]
c) [pic]
13.29: [pic][pic]
13.31: a) [pic]
b) [pic]
13.41: [pic] so for a different acceleration due to gravity [pic]
[pic]
13.43: Besides approximating the pendulum motion as SHM, assume that the angle
is sufficiently small that the length of the spring does not change while swinging in the
arc. Denote the angular frequency of the vertical motion as [pic]
[pic] which is solved for [pic]. But L is...
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