Phy Fluid - Ans.fyi 參考答案

Phy Fluid

awepratization

2010-03-29 5:43 am

1. A wooden cube of side 0.80m and density 700kg/m^3 floats in water of density 1000kg/m^3.
(a) How deep does it go beneath the surface?
(b) What force is needed to push it an extra distance x below the surface?
(c) Show that if the block is pushed down slightly and then released, it will execute SHM. Find the period of oscillation.

2. A hydrometer has a bulb of volume 20cm^3 and a stem of length 40cm and cross-sectional area 0.30cm^2. In water, the hydrometer floats with half the stem immersed. The density of water is 1000kg/m^3.
(a) What is the mass of the hydrometer?
(b) Assume that the scale is marked on the stem only. For what range of densities can it be used?

3. A piece of cork of volume 0.001m^3 and density 300kg/m^3 is kept submerged below the water surface in a tank by means of a thread with its end fixed at the base of the tank. The density of water is 1000kg/m^3. Find the tension in the thread
(a) if the tank is at rest on the ground.
(b) if the tank falls freely.

4. A piece of 2.0kg stone suspended by a thread is completely immersed near the base of a beaker of water. The water level is at a distance of 5.0cm above the base of the beaker. The densities of the stone and water are 3000kg/m^3 and 1000kg/m^3 respectively. Find the extra force on the base of the beaker when the thread is cut off and the stone lies on the bottom of the beaker.

回答 (1)

天同

2010-03-30 5:00 am

✔ 最佳答案

1. (a) 0.8 x 700/1000 m = 0.56 m

(b) Extra upthrust = [(0.8^2).(x).1000g] N, where g is the acceleration due to gravity
Hence, force required = 640gx N

(c) Restoring force = - 640gx N,
hence, -640gx = (0.8^3).(700).a
where a is the acceleration
a = -[640g/(0.8^3).(700)]x = 17.86x
since a is proportional to x, the motion is simple harmonic
Period of oscillation = 2.pi.square-root(1/17.86) s

2. (a) volume below water = [20 + 0.3 x 20] cm^3 = 26 cm^3
Mass of hydrometer = 26 x 1 g = 0.026 kg
(b) Let density of liquid to be measured = d
At lowest density: [20+0.3x40].d = 26
i.e. d = 0.8125 g/cm^3 = 812.5 kg/m3
At highest density: [20].d = 26
i.e. d = 1.3 g/cm3 = 1300 kg/m3

3. (a) Mass of cork = 0.001 x 300 kg = 0.3 kg
Upthrust = 0.001 x 1000 kgf = 1 kgf
Hence, string tension = (1 - 0.3) kgf = 0.7 kgf = 7 N
(b) Let T' be the new tension
T' + mg - upthrust = mg, where m is the mass of the cork
i.e. T' = upthrust = 1 kgf = 10 N

4. Upthrust on stone = (2/3000) x 1000 kgf = 0.667 kgf = 6.67 N
Tension in suspended string = (2 - 0.667) kgf = 1.333 kgf = 13.33 N
Hence, reaction force on stone when it is at the bottom of beaker = 13.33 N

2010-03-29 21:16:01 補充：
sorry...calculation of Q4 is not yet finished.
Extra force on base of beaker = (13.33 - 6.67) N = 6.66 N

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