1. A pump, used to move water through a tube, uses a piston 15 cm in diameter wh

Question

1. A pump, used to move water through a tube, uses a piston 15 cm in diameter which moves at
2.0 cm/s. What is the fluid speed when the tube is 3.0 mm in diameter?


2. What would the apparent weight of a lead brick (2.00 in x 2.00 in x 8.00 in) be if it was
submerged in oil (density of oil; ?o = 0.920 g/cm3 , density of lead; ?L = 11.4 g/cm3)

3. A steel guitar string with diameter of 0.300 mm and a length of 70.0 cm is stretched a length of
0.500 mm while being tuned. How much force is needed to stretch the string by this amount? (
Y = 2.0x1011 Pa)

4. A compressed gas with total mass 203 kg is stored in a spherical container which has a radius of
0.521 m. What is the density of the compressed gas?


5. The Tonga Trench in the Pacific Ocean is 36,000 feet deep. Assuming that sea water has an
average density of 1.04 g/cm3, calculate the absolute pressure at the bottom of the trench in
atmospheres (atm).

6. Calculate the pressure exerted on the ground by a 57.0 kg person standing on one foot. Assume
that the bottom of the person

Explanation / Answer

1.A = pi*r^2 = pi*(0.075m)^2 = 0.01767 m^2

Swept Volume Rate dV/dt = A*v = 0.01767m^2 * 0.02 m/s = 3.53*10^-4 m^3/s

Tube csa is A = pi*r^2 = pi*(.0015m)^2 = 7.069*10^-6 m^2

Fluid velocity is v = dV/dt / A = 3.53*10^-4 m^3/s / 7.069*10^-6 m^2

v = 49.9 m/s

2.2" = 5.08cm. 8" = 20.32cm.
Volume of brick = (5.08 x 5.08 x 20.32) = 521.29cc.
(521.29 x 11.4) = mass of 5,942.7g.
Volume of oil displaced = 521.29cc.
Mass of oil displaced = (521.29 x 0.92) = 479.587g.
Apparent MASS of immersed lead = (5,942.7 - 479.587) = 5,462.413g., or 5.46kg.

3. the diameter:

0.300mm = 0.300mm(1.00m/1,000mm)
= 3.00 x 10??m

The change in length is found the same way and is 5.00 x 10??m. The string

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