Solution to Lab Exercise...051717

SOLVED PROBLEMS #1
1. When 2,100 J is removed from 0.10 kg of a substance, its temperature is observed
    to decrease from 40oC to 10oC.  What is the specific heat of the substance?

    Solution:  Using the formula: Q  =  mcΔT where Q =  2,100 J

2,100 J  =  0.10 kg (c) (40 – 10)oC

c  =  700 J/kg-oC

2. How much heat is required to raise an 800 gm copper pan from 15oC to 90oC if the
    pan contains 1.0 kg of water?

    Solution:  Since both the pan and the water absorb heat, the total heat required is:

Q =  Heat absorbed by the pan  +  Heat absorbed by the water

   =  0.8 kg (386 J/kg-K) (90 – 15)  +  1.0 kg (4185 J/kg-K) (90 – 15)
   =  337,000J

3. In a calorimetry experiment, 0.50 kg of a metal at 100oC is added to 0.50 L of
    water at room temperature in an aluminum calorimeter cup.  The cup has a mass of
    250 gm.  If the final temperature of the mixture is 25oC, what is the specific heat of
    the metal?
    Solution:

In this problem, the calorimer and the water in it will absorb the heat released
by the hot  metal.

0.50 kg (c) (100 – 25)oC = 0.5 kg (4185) (25 – 20) + 0.250 (920) (25 – 20)

c  =  309.67 J/ kg-oC

4. Determine the amount of heat a freezer has to remove from 1.5 kg of water at 22oC                  
    to make ice at – 10oC.

    Solution:
a) Heat required to change the temperature of water at 22oC to 0oC

Q1  =  mcΔT  =  (1.5 kg) (4186 J/kg-oC)(22 – 0)oC
     =  138.14 J

b) Heat required to change water at 0oC to ice at 0oC

Q2  =  mLf  =  (1.5 kg)(3.33 x 105 J/kg)  
                                   =  4.995 x 105 J

c) Heat to be removed to lower ice from 0oC to – 10oC

Q3  = mcΔT  =  (1.5 kg) (2100 J/kg-oC) [0 – (–10)oC]
      = 3.15 x 104 J

Total Heat required:  Qtotal  =  138.14 J +  4.995 x 105 J  +  3.15 x 104 J  

Qtotal  =  5.31 x 105 J

SOLVED PROBLEMS #1
1. When 2,100 J is removed from 0.10 kg of a substance, its temperature is observed
    to decrease from 40oC to 10oC.  What is the specific heat of the substance?

    Solution:  Using the formula: Q  =  mcΔT where Q =  2,100 J

2,100 J  =  0.10 kg (c) (40 – 10)oC

c  =  700 J/kg-oC

2. How much heat is required to raise an 800 gm copper pan from 15oC to 90oC if the
    pan contains 1.0 kg of water?

    Solution:  Since both the pan and the water absorb heat, the total heat required is:

Q =  Heat absorbed by the pan  +  Heat absorbed by the water

   =  0.8 kg (386 J/kg-K) (90 – 15)  +  1.0 kg (4185 J/kg-K) (90 – 15)
   =  337,000J

3. In a calorimetry experiment, 0.50 kg of a metal at 100oC is added to 0.50 L of
    water at room temperature in an aluminum calorimeter cup.  The cup has a mass of
    250 gm.  If the final temperature of the mixture is 25oC, what is the specific heat of
    the metal?
    Solution:

In this problem, the calorimer and the water in it will absorb the heat released
by the hot  metal.

0.50 kg (c) (100 – 25)oC = 0.5 kg (4185) (25 – 20) + 0.250 (920) (25 – 20)

c  =  309.67 J/ kg-oC

4. Determine the amount of heat a freezer has to remove from 1.5 kg of water at 22oC                  
    to make ice at – 10oC.

    Solution:
a) Heat required to change the temperature of water at 22oC to 0oC

Q1  =  mcΔT  =  (1.5 kg) (4186 J/kg-oC)(22 – 0)oC
     =  138.14 J

b) Heat required to change water at 0oC to ice at 0oC

Q2  =  mLf  =  (1.5 kg)(3.33 x 105 J/kg)  
                                   =  4.995 x 105 J

c) Heat to be removed to lower ice from 0oC to – 10oC

Q3  = mcΔT  =  (1.5 kg) (2100 J/kg-oC) [0 – (–10)oC]
      = 3.15 x 104 J

Total Heat required:  Qtotal  =  138.14 J +  4.995 x 105 J  +  3.15 x 104 J  

Qtotal  =  5.31 x 105 J  

Homework , fluids/heat

fluids

7. A 5/8 inch diameter garden hose is used to fill a round swimming pool 6.1 m in

    diameter.  How long will it take to fill the pool to a depth of 1.2 m is water issues

    from the hose at a speed of 0.4 m/sec?

8. A very large vat of water has a hole 1 cm in diameter located at a distance 1.8 m

    below the water level.  How fast does water exit the hose?

9. A house with its own well has a pump in the basement with an output pipe of

    inner radius of 6.3 mm.  Assume that the pump can maintain a gauge pressure of

    410 kPa in the output pipe.  A shower head in the second floor (6.7 m above the

    pump’s output pipe) has 36 holes, each of radius 0.33 mm.  The shower is on “full

    blast” and no other faucet in the house is open.  a) Ignoring viscosity, with what

    speed does water leave the showerhead?  b) With what speed does water move

    through the output of the pump?

10. A water tower supplies water through the plumbing in a house.  A 2.54 cm

    diameter faucet in the house can fill a cylindrical container with a diameter of

    44 cm and a height of 52 cm in 25 sec.  How high above the faucet is the top of the

    tower?  Assume that the diameter of the tower is so large compared to that of the

    faucet that the water at the top of the tower does not move.

heat

7. A hollow aluminum cylinder has an internal capacity of 2.00 liters at 20^oC. It is

    completely filled with turpentine and then slowly warmed to 80^oC.  The average 

    coefficient of linear expansion of aluminum is 24 x 10^–6/^oC and the average

    volume expansion coefficient of turpentine is 9 x 10^–4/^oC)  a) How much 

    turpentine overflows?  b) If the cylinder is then cooled back to 20^oC, will its

    volume be the same?

   

8. What temperature change would cause a 0.10% increase in the volume of a

    quantity of water (β = 2.1 x 10^–4/^oC)  that was initially at room temperature?            

9. A thin spherical shell of silver has an inner radius of 2 x 19^–2 m when the

    temperature is 18^oC.  The shell is heated to 147^oC.  Find the change in the interior

    volume of the shell. The coefficient of volume expansion of shell is 57 x 10^–6/^oC.

10. A brass rod and an aluminum rod of the same diameter are each separately

    attached to an immovable wall opposite each other.  The lengths of the rods are

    2.0 m and 1.0 m, respectively.  When the temperature is 28^oC, the air gap between

    the rods is 1.3 x 10^–3  m.  At what temperature will the gap be closed?

 

Summner 2015 Physics 101 Handouts

Phys 1 - College Physics

Summer 2015

for BSIT/Psyc/Pharma

Click on the link below to download the handouts.

Physics 101AH Lesson1 Handouts

Physics 101AH Lesson2 Handouts

Physics 101AH Lesson 3a Handouts

Physics 101AH Lesson 3b Handouts

Physics 101AH Lesson 4 Handouts

Physics 101AH Lesson 5 Handouts

Midterm Handouts

Physics 101AH: Lesson 5

Physics 101AH: Lesson 5
Work, Power and Energy
Click here to download document

Physics 101AH: Lesson 4

Physics 101AH: Lesson 4

Forces

Click here to download the document

Physics 101 Sample Problems with Solutions - Kinematics

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1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
   

   Given: a = +3.2 m/s2 t = 32.8 s vi = 0 m/s

   Find: d = ??

   d = v_i*t + 0.5*a*t²
   d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s²)*(32.8 s)²
   d = 1720 m
2. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
   

   Given: vi = 18.5 m/s vf = 46.1 m/s t = 2.47 s

   Find: d = ?? a = ??

   a = (Delta v)/t
   a = (46.1 m/s - 18.5 m/s)/(2.47 s)
   a = 11.2 m/s²
   d = v_i*t + 0.5*a*t²
   d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s²)*(2.47 s)²
   d = 45.7 m + 34.1 m
   d = 79.8 m
   (Note: the d can also be calculated using the equation v_f² = v_i² + 2*a*d)
3. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s². Determine the time for the feather to fall to the surface of the moon.
   

   Given: vi = 0 m/s d = -1.40 m a = -1.67 m/s2

   Find: t = ??

   d = v_i*t + 0.5*a*t²
   -1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s²)*(t)²
   -1.40 m = 0+ (-0.835 m/s²)*(t)²
   (-1.40 m)/(-0.835 m/s²) = t²
   1.68 s² = t²
   t = 1.29 s
4. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.
   

   Given: a = -9.8 m/s2 vf = 0 m/s d = 2.62 m

   Find: vi = ??

   v_f² = v_i² + 2*a*d
   (0 m/s)² = v_i² + 2*(-9.8 m/s²)*(2.62 m)
   0 m²/s² = v_i² - 51.35 m²/s²
   51.35 m²/s² = v_i²
   v_i = 7.17 m/s
5. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?
   

   Given: a = -9.8 m/s2 vf = 0 m/s d = 1.29 m

   Find: vi = ?? t = ??

   v_f² = v_i² + 2*a*d
   (0 m/s)² = v_i² + 2*(-9.8 m/s²)*(1.29 m)
   0 m²/s² = v_i² - 25.28 m²/s²
   25.28 m²/s² = v_i²
   v_i = 5.03 m/s
   To find hang time, find the time to the peak and then double it.
   v_f = v_i + a*t
   0 m/s = 5.03 m/s + (-9.8 m/s²)*t_up
   -5.03 m/s = (-9.8 m/s²)*t_up
   (-5.03 m/s)/(-9.8 m/s²) = t_up
   t_up = 0.513 s
   hang time = 1.03 s

PHYSICS 101AH: LESSON 2

PHYSICS 101AH: LESSON 2

SCALARS AND VECTORS

2.1. General Categories of Physical Quantities

Scalar quantities are those quantities which are completely specified by their magnitude,

expressed in some convenient units.

·         easy to handle since they can be manipulated by ordinary laws of algebra (Usually can be added, subtracted, multiplied, divided directly)

·         Examples: length, mass , area, volume, time, density

Vector quantities are those which require for complete specification, both magnitude and

direction.

·         Direction is just as important as the magnitude when specifying quantities

·         It always represented by an arrow.

·         Examples: displacement, force, acceleration, velocity, momentum

·         The sum of the vectors is called the RESULTANT vector.

2.2. Addition of Vectors

            Methods in Adding Vectors

A.     Graphical Method – also known as Geometrical Method and require no computation.

a.       Parallelogram Method (Tail to Tail Method) – used to add only two vectors

b.      Triangle Method (Head to Tail Method) – use to add only two vectors

c.       Polygon Method (Head to Tail Method) – use to add more than 2 vectors

An illustration will help for Head to Tail method: Note that we can move vectors around as we wish on the graph paper, because only the magnitude and direction matter. The location does not. Also note that this is the way we would add together displacements – if I say "go along vector A and then along B", then in the end we would end up in the same place as if we had just gone along vector R, the resultant. In that sense, R is the sum of A and B. We take this as a general definition of adding any two vectors together, whether they are displacements, velocities, or some other quantities.
 
            To do this accurately, we need to use a ruler and protractor. The general technique for adding two vectors on graph paper is as follows:

1.      Start at the origin and draw the first vector (based on the numbers given you)

2.      At the end of the first vector, make a new origin

3.      Draw the second vector starting at the new origin

4.      Connect the tail of the first vector to the head of the second

When you are finished, the last line you have drawn is the resultant. You can then measure the length and angle of this vector using a protractor and ruler.

B.     Analytical Method (Component Method/Trigonometric Method)

Trigonometry is also important in physics. When you have a right-angled triangle, the following relationships are true:

           

This method is based on the fact that we can specify a vector by specifying its magnitude in two perpendicular directions. We take these to be the x and y directions. We call the length of the vector in the x direction as the x-component of the vector, and similarly for y. The nice thing about this method is that once we have the x and y components of the vectors we want to add, adding them is simple. Let us say that we have two vectors, A and B, with the components labeled as A_x, A_y, B_x, and B_y. It should be obvious which component each symbol stands for. If our resultant, or sum vector, is called R, with components R_x and R_y, then we have

R_x = A_x + B_x                                                R_y = A_y + B_y 

But how do we get the components if we are given the angle and direction? We use trigonometry. Consider the diagram:

We know from trigonometry that
                        A_x = A cos q 
                        A_y = A sin q

Similar relations hold for the components of B, or any other vector for that matter. Here A is the magnitude of A, and q is the angle.

We now need to be able to go back from the components to the magnitude and direction. We have, again from trigonometry,
                                    R² = R_x² + R_y² 
                                    q = tan^-1 (R_y / R_x)  

 The process for adding two vectors, A and B, is thus:

1. Find the components of A and B using (2a) and (2b)
2. Add the components together using (1a) and (1b)
3. Find the magnitude and direction of R using (3a) and (3b)

Note: Calculators are funny things. You should always make a sketch of your addition to see if your values agree with what your calculator gives. The magnitudes should be correct, but the angles will often come out wrong. To adjust the angles, note that

                        cos q = cos (360° - q )

                         sin q = sin (180° - q )

                         tan q = tan (180° + q )

            The summarized steps for component method are as follows:

1.      Resolve all the given vectors into their x and y components.

2.      Find the algebraic sum of all the x-components (∑X), and the algebraic sum of all the y-components (∑Y).

3.      Find the magnitude of the resultant using the Pythagorean theorem.

R = sqrt ((∑X)² + (∑Y)²)

4.      determine the angle of the resultant and the specific direction using the tangent function.                 

θ = arctan (∑Y /  ∑X)