Physics 1LIIA Week 9[2107] PDF

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Weekly Schedule 1LII Week N°9, 8 - 12 Nov. Consonni Physics

Teacher Discipline

General Theme

•

Equilibrium in fluids

Relevant points on the topics

•

Stevin’s Law (hydrostatic pressure)

Notes Theory → Study: pages 139-142. Exercises → Page 155 n 31, 36, 38, 41, 42 + review exercises.

Please note that relevant topics also represent possible questions, together with the related exercises. Keywords are in bold letters.

Keyword

Parola chiave

Definition

Stevin’s law or hydrostatic pressure law

Legge di Stevino

The pressure difference (𝑝2 − 𝑝1 ) between two points in a column of fluid at equilibrium is given by the density of the fluid (d) times the g-constant (g) times the difference in height between the two points (ℎ1 − ℎ1 ).

Communicating vessels

Vasi comunicanti

Containers joined by some system well below the surface of the fluid, so that the fluid can move from one container to the other.

Communicating vessels law

Principio dei vasi comunicanti

In a system of communicating vessels, the height reached by the liquids (h1 and h2) are inversely proportional to their densities (d1 and d2).

Hydrostatic pressure exercises Page 154 #31 The picture shows a section of a lake. Points A, B and C are at the same depth, and so are D, E and F. Pressures in A, D and G measure: pA = 1,6 ∙ 105 Pa, pD = 2,2 ∙ 105 Pa; pG = 2,9 ∙ 105 Pa. What is the water pressure in B, C, E, F?

[1,6 ∙ 105 Pa; 1,6 ∙ 105 Pa; 2,2 ∙ 105; 2,2 ∙ 105] Page 155 #36 A container is partially filled with water and it is covered with a cork. A part of the air on top of the liquid has been sucked out. The picture represents the graphic of pressure p in the liquid depending on depth h. -

Determine the pressure of the air left in the container. Determine the liquid’s density.

[1,2 ∙ 103 Pa; 610 kg/m3]

Page 155 #38

The graphic shows the trend of pressure (in 105 Pa units) in a liquid as a function of depth h (in meters). On the free surface of the liquid, i.e. at h = 0 m, there is only atmospheric pressure.

-

Determine the liquid’s density. Which liquid can it be? [13600 kg/m3]

Page 156 #41 In a container, open at the top, a certain quantity of water (density 1000 kg/m3) and a certain quantity of oil (density 800 kg/m3) are poured. The two liquids don’t mix. On the oil’s free surface we have only atmospheric pressure. Point A is at a depth hA = 0,500 m and point B is at a depth hB = 1,50 m. The oil column has a height hO = 1,00 m. -

Compute the pressure in A. Compute the pressure in B.

Page 157 #42 A U tube is filled with water (density 1000 kg/m3). In one of the two branches some oil is added (density 800 kg/m3). The picture shows the two liquids at equilibrium Point B is on the separation surface between oil and water. Point A is at the same height as B. The height of the oil column is hO = 15 cm. Both branches are open at the top. Find ha, height of the water column above point A. [12 cm]

Review and consolidation exercises (optional) 1. Write the following using scientific notation (solutions in the next page): • 1473000000000 • 0,0000017005 2. Simplify the following expression (solutions in the next page): 21 ∙ 107 ∙ 4 ∙ 10-5 14 ∙ 10-4 ∙ 2 ∙ 10-3 3. Rearrange the following equations to find the requested quantity (solutions in the next page). • p = m v, find m. •

P =

•

E =

I2

, find R.

R F , q

find F.

4. Draw the force-space graph of a spring, knowing that if you apply on it a force F=4 N, the elongation of the spring is x=1 cm (solution in the next page). 5. Lucy and Julia are in front of each other and pull on two force meters attached to the same object. The force meters measure the force with which they are pulling on the object. Lucy’s force meter reads 1,4 N, Julia’s reads 2,2 N. What is the modulus of the total force on the object? And its direction? [0,8 N from the object to Julia] 6. In a hydraulic lift a force of modulus F = 35 N applied to the piston of smallest area balances out a weight P = 1750 N on the piston of largest area. Determine the ratio between the area of the largest piston and the area of the smallest piston. [50 N]

Review and consolidation exercises, solutions (optional) 1. • •

1473000000000 = 1,473 · 1012 0,0000017005 = 1,7005 · 10-6

2. 21 ∙ 107 ∙ 4 ∙ 10-5 12 ∙ 10-4 ∙ 2 ∙ 10-3

= 7 ∙ 109

3. •

p = m v, find m. m =

•

P =

•

E = , find F. F = qE q

I2 R F

, find R. R =

I2 P

4.

5. [0,8 N from the object to Julia] 6. [50 N]

p v...