Ch. 14 WP Reading

When there is greater pressure behind an element, what happens to the acceleration?

Direction is forward; the pressure difference produces an acceleration in the forward direction and is the reason the element's speed increases.

The U-tube contains 2 liquids in static equilibrium: Water of density pw (=998 kg/m3) is in the R arm, and oil of unknown density px is in the left. Measurement gives l = 135 mm and d = 12.3 mm. What is the density of the oil?

Equations are equal because the water is not moving; must be balanced.

What happens to kinetic energy density as pressure increases?

Kinetic energy density increases as pressure decreases

In a fIn a fluid stream, which is true about a region in which the streamlines are relatively close?

Relatively low pressure The flow speed there is relatively large and thus the pressure is relatively low.

Ethanol of density p = 791 kg/m3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A1 = 1.20 x 10^-3 m2 to A2 = A1/2. The pressure difference between the wide and narrow sections of pipe is 4120 Pa. What is the volume flow rate Rv of the ethanol?

Rv = v1A1 = v2A2, but since we have 2 unknown speeds, we need to use Bernoulli's equation, using the fact that A2 = A1/2: v1 = Rv/A1 v2 = Rv/A2 = 2Rv/A1 Substitue into Bernoulli's equation (p1+1/2..=p2+1/2..), solve for Rv

The figure shows four containers of olive oil. Rank them according to the pressure at the depth h. Indicate the ranking by moving the labels into the boxes, with the greatest pressure at the top left. Use side-by-side boxes for any ties, in alphabetic order.

Shape of container is irrelevant, only depth matters.

Why is work negative in Bernoulli's Equation?

The upward displacement and the downward gravitational force have opposite directions.

Here are three situations in which a force is uniformly applied to a flat surface. The force magnitudes and surfaces areas are given. Rank the situations according to the pressure on the surface. Indicate the ranking by moving the labels into the boxes, with the greatest pressure at the top left. Use side-by-side boxes for any ties, in numeric order.

Uniform: p = F/a

The figure shows a stream of water emerging from a faucet "neck down" as it falls. This change in the horizontal cross-sectional area is characteristic of any laminar (non-turbulent) falling stream because the gravitational force increases the speed of the stream. Here the indicated cross sectional areas are Ao = 1.2 cm2 and A = 0.35 cm2. The two levels are separated by a vertical distance h = 45 mm. What is the volume flow rate from the tap? b) In the video we found that the volume flow rate from the tap is RV= 34 cm3/s. What is the value of RV after the water has fallen by the given distance h= 45 mm?

Volume flow rate through higher cross section must be sam as that through lower cross section. Aovo = Av, where vo and v are water speeds at levels corresponding to Ao and A. Since water is free falling w/ acceleration g, can also use v²= v²₀ +2gh. Eliminate v by solving Aovo = Av for v, and plugging new v equation into free fall equation. you get vo = √(2ghA²)/A²₀-A²), plug in numbers. The odd values should cancel out, no need to convert. b) Rv = Aovo

A person with a mass of 70 kg floats in water. What is the person's apparent weight?

Zero. The magnitude of the buoyant force on a floating body is equal to the body's weight. Thus a floating body has an apparent weight of zero—the body would produce a reading of zero on a scale.

A penguin floats first in fluid 1 of density ρ0, then in fluid 2 of density of 0.95ρ0, and then in fluid 3 of density 1.1ρ0. a) Rank the fluids according to the magnitude of the buoyant force on the penguin. b) Next, rank the fluids according to the amount of fluid displaced by the penguin.

a) In each fluid, floating requires that the buoyant force match the penguin's weight, which (of course) does not change. b) 2, 1, 3 (less density, more displacement). Recall that the buoyant force is equal to the weight of the displaced fluid. Also recall that if the penguin is to float, then the buoyant force must match its weight. So, more of the low-density fluid must be displaced in order to get enough buoyant force to match the penguin's weight.

A living room floor has dimensions 3.5 m, 4.2m, and 2.4 m. a) What is the mass of the air when density = 1.0 atm? b) What is the atmospheric Force downwards on top of head when head volume = 0.040 m^2?

a) p = density p = m/v, m=pv, include g for gravity so mg = (pv)g b) When considering uniform pressure: p = F/a, F = pa. Converting atm to N: 1.0 atm x (1.0 x 10⁵ N/m²/1atm) = 4.0 x 10³ N

We will release five spheres fully submerged in water (density of 1000 kg/m3) to see how they accelerate in a vertical race. Their masses and volumes are given here.

p = m/v. Those with greater density than water will sink, with greatest densities sinking fastest. Those with lower densities than water will rise, with the least densities rising fastest.

Gauge pressure on a scuba diver exhaling during descent; When diver reaches surface from depth L, the difference between external pressure on him and air pressure in his lungs is 9.3 kPa (Pa = Pascal). From what depth dos he start?

p = p0 + pgh, substitute L for h. To find L, use delta p = p - p0 = pgL L = delta p/ pg = 9300 Pa / (998)*(9.8) Water density would have been given here.

How to calculate the acceleration of an object ascending towards the surface fo a liquid before it reaches the surface, in terms of ratio a/g.

pf/p-1

Bernoulli principle for a leaky tank: In the old West, a desperado fires a bullet into an open water tank, creating a hole distance h below the water surface. What is the speed v of the water exiting the tank?

v =√2gh