Download presentation

Presentation is loading. Please wait.

1
**March 28, 2011 DRILL Solve these problems**

U3f – L4 March 28, 2011 DRILL Solve these problems A teacher standing on a 6in by 8in block of wood exerts a pressure of 4 psi. How much does the teacher weigh? A basketball originally has 10 psi of pressure and a volume of 50 mL. If the volume is decreased to 10 mL, what is the resulting pressure? Take out handout from Friday. We will talk about the solutions in 10 minutes.

2
U3f – L4 Drill #1 A teacher standing on a 6in by 8in block of wood exerts a pressure of 4 psi. How much does the teacher weigh? Step 1: Write given information P = 4 psi F = ?? A = ? Step 2: Write the formula and solve A = L*W=6 in *8 in = 48 in2 P = F/A 4psi = F/48in2 F = 4psi *48in2 F=192 lbs

3
U3f – L4 Drill #2 A basketball originally has 10 psi of pressure and a volume of 50 mL. If the volume is decreased to 10 mL, what is the resulting pressure? Step 1: Write given information P1 = 10psi V1 = 50 mL P2 = ?? V2 = 10mL Step 2: Write the formula and solve P1*V1=P2*V2 10psi * 50 mL = P2*10mL 500 psi*mL = P2*10mL P2 =500 psi*mL/(10mL) P2=50psi

4
**Worksheet #4 P1V1 = P2V2 (80 psi) x (6 L) = P2 x (10 L)**

U3f – L4 Worksheet #4 A gas system has initial pressure and volume of 80 psi and 6 L. If the volume changes to 10 L, what will the resultant pressure be in psi? Step 1: Write given information P1 = 80 psi V1 = 6 L P2 = ? V2 = 10 L Step 2: Write the formula and solve P1V1 = P2V2 (80 psi) x (6 L) = P2 x (10 L) 480 psi*L = P2 x (10 L) 10 L L P2 = 48 psi

5
**Worksheet #5 P1V1 = P2V2 P1 x (300 cm3) = 5 psi x (2.5 cm3)**

U3f – L4 Worksheet #5 A sample of hydrogen gas is compressed from 300 cm3 to 2.5 cm3. Its pressure is now 5 psi. What was the original pressure of the helium? Step 1: Write given information P1 = ? V1 = 300 cm3 P2 = 5 psi V2 = 2.5 cm3 Step 2: Write the formula and solve P1V1 = P2V2 P1 x (300 cm3) = 5 psi x (2.5 cm3) P1 x (300 cm3) = 12.5 psi*cm3 300 cm cm3 P1 = psi

6
**Worksheet #6 P=F/A P=50lbs/0.75 ft2 P=66.7lbs/ft2**

U3f – L3 Worksheet #6 The surface of a concrete block that touches the ground is 0.75 ft2. What is the pressure of the concrete block on the surface of the ground if the block weighs 50 lbs. Step 1: Write given information P = ? A = 0.75 ft2 F = 50 lbs Step 2: Write the formula and solve P=F/A P=50lbs/0.75 ft2 P=66.7lbs/ft2

7
U3f – L3 Worksheet #7 The diagram below illustrates how hydraulic brakes in a car work. The pedal must be pressed with a force of 8 lbs. The surface area of the piston connected to the pedal is 1 square inch. If the surface area of the piston connected to the other end of the brake line is 1.5 square inch, what is the force applied to that piston? P1 = P2 P1 = F1 / A1 P2 = F2 / A2

8
**Solution: Worksheet #7 Step 1: Write given information P1 = ? P2 = ?**

U3f – L3 Worksheet #7 Solution: Step 1: Write given information P1 = ? P2 = ? F1 = 8 lbs. F2 = ? A1 = 1 in2 A2 = 1.5 in2 Step 2: Write the formula and solve for the unknowns P1 = F1 / A1 P1 = (8 lbs) / (1 in2) P1 = 8 psi P1 = P2 = 8 psi F2 = 8 psi x 1.5 in2 = lbs. P2 = F2 / A2 F2 = P2 x A2

9
**Worksheet #8 P=F/A 85psi=F/0.035 in2 85psi*0.035 in2 =F**

U3f – L4 Worksheet #8 A bicycle tire is holding air at 85 psi of pressure. The valve of the tube lets air in and out. If the cross sectional area of the valve is in2, how much force is required to add more air to the tube? Step 1: Write given information P = 85 psi A = in2 F = ? Step 2: Write the formula and solve P=F/A 85psi=F/0.035 in2 85psi*0.035 in2 =F F =2.975 lbs

10
**Worksheet #9 P1=P2 P1 = F1 / A1 P2 = F2 / A2**

U3f – L4 Worksheet #9 A bicycle pump can be modeled like the diagram in number 7 above. Suppose the tube has 85 psi of pressure inside. To add more air to the tube, how much force is required to push the handle end of a bike pump if its cross-sectional area is 1 in2 P1 = F1 / A1 P2 = F2 / A2 P1=P2

11
**Solution: Worksheet #9 Step 1: Write given information**

U3f – L4 Worksheet #9 Solution: Step 1: Write given information P1 = 85psi P2 = 85psi F1 = lbs. F2 = ? A1 = in2 A2 = 1 in2 Step 2: Write the formula and solve for the unknowns P2 = F2 / A2 85psi= F2 / (1 in2) F2= 85 psi* 1 in2 F2 = 85 lbs

12
**Worksheet #10 P=F/A 85psi=170lbs/A 85psi*A=170 lbs**

U3f – L4 Worksheet #10 If we doubled the amount of force used in number 9 above, what must we change the area of the pump to be in order to maintain the same pressure Step 1: Write given information P = 85 psi A = ??? F = 170 lbs Step 2: Write the formula and solve P=F/A 85psi=170lbs/A 85psi*A=170 lbs A=170lbs/85 psi A=2in2

13
**Quiz on Wednesday or Thursday. Understand the problems**

U3f – L4 HOMEWORK Quiz on Wednesday or Thursday. Know the definitions. Understand the problems

Similar presentations

OK

Pressure Worksheet 1. I push on the handle of a bicycle tire pump with a forceof 100 Newtons . The pist on has an area of 0.001 m 2. What is the pressure.

Pressure Worksheet 1. I push on the handle of a bicycle tire pump with a forceof 100 Newtons . The pist on has an area of 0.001 m 2. What is the pressure.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Download ppt on mind controlled robotic arms nasa Ppt on lymphatic system Ppt on you can win if you want Ppt on porter's five forces template Ppt on diode as rectifier circuit Ppt on national education day of bangladesh Ppt on remote controlled screw jack Ppt on adjectives for class 2 Ppt on electric meter testing jobs Ppt on motivation for teachers