Hydrostatic pressure varies only with the vertical distance and is independent of the shape of the container. Thus, the pressure increases with depth in fluid. In this experiment we will investigate the resultant hydrostatic force applied on a submerged surface (the rectangular surface of Toroid ). Plus, determining the center of pressure experimentally and theoretically.

Experimental Procedures:

1. Leveling the Flotation Tank
2. Adjusting the counterbalance weight. (leveling the balance arm) 3. Initial filling of the floatation tank.
4. Adding weight.
5. Filling the tank to rebalance the added weight. (re leveling the balance arm) 6. Measuring the water level.
7. Draining the...

...Surface Tension of Liquids
Karen Mae L. Fernan
Department of Chemistry, Xavier University-Ateneo de Cagayan, Philippines
Date performed: Nov. 22, 2012 ∙ Date Submitted: January 16, 2013
E-mail: fernankarenmae26@yahoo.com
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Abstract
Surface tension is defined as the energy or work required to increase thesurface area of a liquid due to intermolecular forces. Measuring the surface tension is useful in identifying unknown liquids and to check for its purity. By capillary method, the surface tension of a given unknown liquid was obtained. Water was used as reference liquid and both liquids were subjected to capillary method at 20, 30 and 40℃. The surface tension of water was used to calculate the radius of the capillary. The equation γ= 12ρgrh , where g is the acceleration due to gravity, r is the calculated radius of the capillary at a certain temperature, h is the capillary rise and ρ is the density of the liquid, was used to calculate for the surface tension of the unknown liquid. In the experiment, the calculated surface tension of the unknown liquid H was 22 dyne/cm at 20℃ , 22 dyne/cm at 30℃ and 18 dyne/cm at 40℃. It was found to be close to the surface tension value of isopropyl alcohol...

...1. Which equation below represents the quadratic formula?
*a. -b±b2-4ac2a = x
b. a2+b2=c2
c. fx=a0+n=1∞ancosnπxL+bnsinnπxL
2. Which of the following represents a set of parallel lines?
a. Option one
b. Option two
*c. Option three
3. What is the definition of an obtuse angle?
*a. an angle greater than 90°
b. an angle equal to 90°
c. an angle less than 90°
4. Which formula below represents the area of a circle?
a. A=2πr
*b. A=πr2
c. A=π2r
d. A= √π
5.
What geometric term is represented by the image below?
a. a corner
*b. a cross-section
c. the circumference
d. the perimeter
11. Using the data in the table below, calculate the mean, or average, number of points scored by Player B.
| Game 1 | Game 2 | Game 3 | Game 4 | Game 5 |
Player A | 13 | 12 | 9 | 11 | 13 |
Player B | 12 | 11 | 15 | 20 | 12 |
*a. 14
b. 11.5
c. 13
d. 13.67
6. This instrument is commonly used by surveyors. It measures horizontal and vertical angles to determine the location of a point from other known points at either end of a fixed baseline, rather than measuring distances to the point directly. What is it called?
a. triangulator
b. binocular
c. tripod
*d. theodolite
7. What is the name of the missing shape in the flowchart below?
a. Acute
b. Obtuse
*c. Isosceles
d. Right
8. What category includes all of the items on the list below?
* Square
* Rectangle
*...

...Results
Part A: Rectangular Notch
Width of rectangular notch, B = 3.00 cm
Volume (L)
Height, H (m)
Time, t (s)
3.00
0.01
26.73
3.00
0.02
14.76
3.00
0.03
8.91
3.00
0.04
5.40
3.00
0.05
3.55
3.00
0.06
3.01
Part B: Vee Notch
Enclosed angle of Vee notch, θ = 90°
Volume (L)
Height, H (m)
Time, t (s)
3.00
0.01
56.20
3.00
0.02
15.84
3.00
0.03
7.56
3.00
0.04
3.64
Calculations
Part A: Rectangular Notch
Volume Collected (m3)
Height of Water Level, H (m)
Time for Collection, t (s)
Volume Flow Rate, Qt (m3/s)
H3/2 Rectangular Notch (m3/2)
Rectangular Notch Discharge Coefficient, Cd
0.003
0.010
26.73
1.12×10-4
1.00 × 10-3
1.264
0.003
0.020
14.76
2.03×10-4
2.83 × 10-3
0.810
0.003
0.030
8.91
3.37×10-4
5.20 × 10-3
0.732
0.003
0.040
5.40
5.56×10-4
8.00 × 10-3
0.785
0.003
0.050
3.55
8.45×10-4
1.12 × 10-2
0.852
0.003
0.060
3.01
9.97×10-4
1.47 × 10-2
0.766
For first reading:
1. Volume Flow Rate, Qt
2. Rectangular Notch, H3/2
3. Discharge Coefficient, Cd
4. Percentage Error
From the theory, the Cd value is 0.6
From the table, one value of Cd is taken which is, Cd = 0.732
Part B: V-Notch
Volume Collected (m3)
Height of Water Level, H (m)
Time for Collection, t (s)
Volume Flow Rate, Qt (m3/s)
H5/2 Rectangular Notch (m3/2)
Vee Notch Discharge Coefficient, Cd
0.003
0.01
56.20
5.34×10-5
1.00 × 10-5
2.260
0.003
0.02
15.84
1.89×10-4
5.66 × 10-5
1.414
0.003
0.03
7.56
3.97×10-4
1.56 × 10-4
1.077
0.003
0.04
3.64
8.24×10-4
3.20 × 10-4
1.090...

...on a film. Wetting, in simple terms, is the ability of a liquid to spread across the surface of a
solid to produce a uniform, continuous surface. How a solid is settled by a liquid is measured by the surface tension of the liquid relative to the surface tension of the solid. For a coating Adhesive or ink to be usable it must a) wet and adhere to the substrate, b) cure, and c) exhibit excellent enduse properties. Thesurface tension of a solid has important influence on these properties. A solid surface with intimate coverage of a liquid is necessary to produce a strong uniform adhesive, coating or ink bond.
Surface tension is a property of the surface of a liquid that allows it to resist an external force. It is revealed, for example, in floating of some objects on the surface of water, even though they are denser than water, and in the ability of some insects (e.g. water striders) to run on the water surface. This property is caused by cohesion of similar molecules, and is responsible for many of the behaviors of liquids.
Surface tension has the dimension of force per unit length, or of energy per unit area. The two are equivalent—but when referring to energy per unit of area, people use the term surface energy—which is a more general term in the sense that it applies also to solids and not just...

...
2 CALCULATIONS
For the sample calculations, we looked at the first sample point of the flow in Pipe 1, the smallest diameter smooth copper tube:
The first step in determining the properties of the flow is finding the density and kinematic viscosity of the water. At 296.51 K, water has the following properties1:
From this we can determine the bulk velocity of the stream using Equation 1.
(Eqn. 1)
Where is the flowrate in m3/s and A is the cross-sectional area of the pipe. To find the flowrate, we multiply the flowmeter reading by the constant
and convert from gallons to cubic meters as follows:
The cross sectional area of the 7.75mm pipe is
Plugging these values into Equation 1, we obtain a bulk velocity .
With the bulk velocity value, we can find the Reynolds number of the flow using Equation 2.
(Eqn. 2)
Plugging in known values to Equation 2, we find:
The experimental friction factor of the pipe can be calculated as:
(Eqn. 3)
Using the pressure drop for the chosen sample from smallest smooth copper pipe across the known distance L, we obtain an experimental friction factor
The theoretical friction factor for smooth pipes can be calculated with the Petukhov formula:
(Petukhov Formula)
Using this formula with our calculated Reynolds number yields a theoretical friction factor of
Because Pipe 4 is a rough pipe, this...

... CHEMISTRY 2
SURFACE TENSION
AIM
To find the surface tension of different liquids.
INTRODUCTION
Surface tension is a contractive tendency of the surface of a liquid that allows it to resist an external force. Surface tension is an important property that markedly influences the ecosystem. Surface tension is exposed, for example, any time an object or insect (e.g. water striders) that is denser than water is able to float or run along the water surface. At liquid-air interfaces, surface tension results from the greater attraction of water molecules to each other (due to cohesion) than to air (due to adhesion). The net effect is an inward force at its surface that causes water to behave as if its surface were covered with a stretched elastic membrane. Because of the relatively high attraction of water molecules for each other, water has a high surface tension (72.8 millinewtons per meter at 20°C) compared to that of most other liquids. Surface tension is an important factor in the phenomenon of capillarity.
Surface tension has the dimension of force per unit length, or of energy per unit area. The two are equivalent—but when referring to energy per unit of area, people use the term surface energy—which is a more general...

...CALCULATION
Before starting on any hydro power generation project it is important to survey the proposed site to calculate the amount of available hydro power.
The two important factors to consider are the flow and the head of the stream or river. The flow is the volume of water which can be captured and re-directed to turn the turbine generator, and the head is the distance the water will fall on its way to the generator. The larger the flow - i.e. the more water there is, and the higher the head - i.e. the higher the distance the water falls - the more energy is available for conversion to electricity. Double the flow and double the power, double the head and double the power again.
A low head site has a head of below 10 metres. In this case you need to have a good volume of water flow if you are to generate much electricity. A high head site has a head of above 20 metres. In this case you can get away with not having a large flow of water; because gravity will give what you have an energy boost.
The key equation to remember is the following:
Power = Head x Flow x Gravity
where; Power is measured in Watts
Head in metres
Flow in litres per second
Acceleration due to gravity in metres per second square
The acceleration due to gravity is approximately 9.81 metres per second per second - i.e. each second an object is falling, its speed increases by 9.81 metres per second (until it hits its terminal velocity).
Example:
Head = 36 m ;...

...Summer 2010-3 CLASS NOTES CHAPTER 1
Section 1.1: Linear Equations
Learning Objectives:
1. Solve a linear equation
2. Solve equations that lead to linear equations
3. Solve applied problems involving linear equations
Examples:
1. [pic]
[pic]
3. A total of $51,000 is to be invested, some in bonds and some in certificates of deposit (CDs). If the amount invested in bonds is to exceed that in CDs by $3,000, how much will be invested in each type of investment?
4. Shannon, who is paid time-and-a-half for hours worked in excess of 40 hours, had gross weekly wages of $608 for 56 hours worked. What is her regular hourly wage?
Answers: 1. [pic]
2. [pic]
3. $24,000 in CDs, $27,000 in bonds 4. $9.50/hour
Section 1.2: Quadratic Equations
Learning Objectives:
1. Solve a quadratic equation by (a) factoring, (b) completing the square, (c) the
quadratic formula
2. Solve applied problems involving quadratic equations
Examples:
1. Find the real solutions by factoring: [pic]
2. Find the real solutions by using the square root method: [pic]
3. Find the real solutions by completing the square: [pic]
4. Find the real solutions by using the quadratic formula: [pic]
5. A ball is thrown vertically upward from the top of a...