help with the physic

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Do the lap and put the data from you for all three laps

This is one of it https://morgan.blackboard.com/bbcswebdav/pid-19713... and I attached the rest an

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The Acceleration Due to Gravity Prof. M.L.C. Murdock version: Sept2015 Introduction To sensibly describe the motion of objects in our universe, we need to understand displacement, velocity and acceleration. This lab will emphasize the latter. When we use the word “acceleration” we mean the rate at which the velocity of a moving object changes with time. Accelerations are always caused by “forces.” (For those who missed lecture, think of a force as a push or a pull.) This is the essence of Newton’s first law. In today’s lab we will measure the acceleration due to the gravitational force exerted by the earth on two different types of objects, a tennis ball and a ping-pong ball. Theory Newton's second law of force relates the amount of force on an object to its mass and acceleration. F = ma (1) The greater the force on an object, the larger is its acceleration. Beyond that, this equation doesn’t say all that much until we know what to write on the left-hand side! Fortunately, that is the case for today. Probably the most apparent (and yet the weakest!) of the known forces in nature is the gravitational force. Newton's Universal Law of Gravitation describes the mutual attractive gravitational force (Fg) exerted on each other by two objects with masses m and m’ that are separated by a distance r. The magnitude of this force is simply Fg = G m(1) m(2) (2) d^2 The constant G is called the gravitational constant and, as its name implies, does not depend on variables like mass or distance. This course will not explain where this formula comes from or even the experimental tests done to verify it. Instead, we will just use it. To compute the magnitude of the gravitational force between the earth and an object, we substitute the mass of the earth (ME) and the distance from the object to the center of the earth (r). When the objects are on or near the earth's surface, this distance can be approximated by the value for the radius of the earth (RE), so that Equation (2) becomes Fg = (3) We see that the force on the object depends only on the mass of the object, because G, ME, and RE are all constants and remain the same if we change our object. This force (measured at the earth's surface) is called the weight of the object. Looking at Equation (1) and equating F to the gravitational force (Fg), we see that: ma = = mg , (4) where a= . This quantity g is a special acceleration and is itself a constant because it depends on quantities that do not change with time. We call this special acceleration the gravitational acceleration and is, for example, the same acceleration a book experiences when you drop it. Allegedly, Galileo first demonstrated this result when he dropped cannonballs of different masses (weights) from the Leaning Tower of Pisa to show that although they had different masses, when dropped together, they landed together. This happened in this manner because they both experienced the same acceleration. A similar experiment can be done by dropping a coin and a feather. When dropped in air, the coin always lands first, but when they are dropped in a vacuum, an environment where there is no air, they land together! In the coin and feather case, the different velocities are due to the presence of a force, the frictional force on the coin and the feather due to the presence of air. Our Equation (4) equates the total force to the gravitational force and therefore neglects the effects of air friction. Let’s now try to discover a quantitative method for determining the gravitational acceleration, g. We first look at the equation for displacement x in a one dimensional universe: x(t) = vo t + (1/2) at 2 (5) Here, the quantities vo and a are, respectively, the initial velocity and the acceleration. In this experiment, like Galileo, we will be dropping an object from rest, so that vo will be zero. We will assume that the only acceleration present is due to the force of gravity so that we can set a = g. We then obtain the relation that describes the distance the object falls as a function of time x(t) = (1/2) gt2 . (6) Equation (6) provides a means to measure g. All we need to do is to drop an object through a known distance x and then measure the time t it takes to fall. If we know both x and t, we can solve equation (6) for g, g = (7) Hence, by measuring the travel distance and the travel time for an object falling from rest, we can measure the gravitational acceleration. Procedure Data Use the given lab form to record all data and extended data (t 2 ). You should also record m, the mass of the ball. Remember to report all data with units. For the last column of all four tables, the value of g to be used in computing |gave – g| is your value of g found from that particular drop. Calculations 2 1. Using the measured values for x and the calculated values for t , insert these values into Equation (7) to get the measured values for g. Show all calculations and remember to label the appropriate quantities with their respective units. 2. Find the average of the five measurements of g and record this in the table below. 3. As a measure of precision, we will be using the average deviation from the mean for the measured value of the gravitational acceleration. Reread the lab write-up for Measurement and Measurement Error to remind yourself on how to do this type of error analysis. Your final result for the gravitational acceleration should be reported as: g = gave ±  As a measure of accuracy, use the given accepted value for g (9.8 m/s2) and your measured average value of g to compute the percentage error. Again, reread the Measurement and Measurement Error write-up to refresh your memory on how to do this. Conclusion 1. How did the gravitational acceleration vary with the different heights? Did it become smaller, larger or remain roughly constant? Why do you think this is? 2. What did you discover in this experiment? Error Analysis 1. 2. How well do you think you can measure distance in this lab? Explain. How well do you think you can measure drop time in this lab? Is this a matter of accuracy, precision or both? Explain. 3. Examine equation (7). Do you think it is more important to measure accurately the drop time or the fall distance of the ball to determine g with the greatest possible accuracy? Explain. 4. Compare your value of g with the accepted value of 9.8 m/s2. Is your value lower or higher than this value? Why do you think this is? Try to identify the experimental conditions or errors that may cause this discrepancy. Ball type: ______________________ trial 1 X t t2 g |gave-g | 2 3 4  Ball mass Average measured value for g= _______________ Accepted value for g= Calculations _______ % error: _______ Conclusions 1. 2. 3. Error Analysis 1. 2. 3. 4. Physics 101 DIY : Make and Use a Barometer to Measure Air Pressure Overview Air pressure is the result of the weight of tiny particles of air (air molecules) pushing down on an area. While invisible to the naked eye (i.e. microscopic), they nevertheless take up space and have weight. For example, take a deep breath while holding your hand on your ribs and observe what happens. Did you feel your chest expand? Why did it expand? Air pressure expands because the air molecules take up space in your lungs, causing your chest to expand. Furthermore, air can be compressed to fit in a smaller volume since there's a lot of empty space between the air molecules. When compressed, air is placed under high pressure. Meteorologists measure these changes in the air to forecast weather, and the tool they use is a barometer. The common units of measurement that barometers use are millibars (mb) or inches of mercury. DIY Make a Barometer A. Materials o B. Theory How does this measure air pressure? C. Procedure 1. Place the completed barometer and scale in a shaded location free from temperature changes (i.e. not near a window as sunlight will adversely affect the barometer's results). 2. In your notebook or the table below, record the current date, time, the weather conditions, and air pressure (i.e. the level where the end of the straw measures on the scale). 3. Continue checking the barometer twice a day (if possible) each day over a four days period. Date Data Table Weather Air Time Conditions Pressure Date June 4, 2003 June 4, 2003 June 5, 2003 Sample Data Table Weather Air Time Conditions Pressure Clear and 9:30 am 4 Sunny 2:30 pm Cloudy 3 9:30 am Rainy 1 Barometer Analysis Answer the following questions in complete sentences. 1. What problem were you trying to solve with your barometer? ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ 2. Were there any changes in the weather during the week? ___________________________________________________________________________ ___________________________________________________________________________ 3. Did the barometer measurement change when the weather changed? How much? ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ 4. Did the barometer measurement change without a change in weather? Why do you think that happened? ___________________________________________________________________________ ___________________________________________________________________________ 5. How well did your barometer work? ___________________________________________________________________________ ___________________________________________________________________________ 6. What would you change if you could design the barometer again? ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ 7. How do the barometer measurements help us understand the system of weather around us? ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________ ___________________________________________________________________________
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Explanation & Answer

here are the 2 complete labs

Exercise 1:
It is given that the car travels 20 miles at 600 north of west, and then 35 miles at 45o north of east. Let

A and B be the vectors that represent the 1st and 2nd displacements, respectively. By letting the +xaxis due east and the +y-axis due north, we have each displacement vector expressed in the unit vector
notation as follow:

1
3
A  20cos  60o  x  20sin  60o  y  20  x  20 
y  10x  10 3y
2
2
2
2
35 2
35 2
B  35cos  45o  x  35sin  45o  y  35 
x  35 
y
x
y
2
2
2
2
The resultant displacement vector in unit vector notation can be obtained using the component method
as follow:

R=A+B
 35 2
35 2 
 10x  10 3y  
x
y 
2
 2



35 2 
35 2 
  10 
 x  10 3 
y
2 
2 







Hence, the magnitude and direction of the resultant displacement vector can be calculated by:
2

R  Rx  Ry
2

 Ry
 Rx

  tan 1 

2

2


35 2  
35 2 
  10 
  10 3 
  44.58
2  
2 




35 2  
35 2  
1
o
  10 3 
   70.68
  tan   10 
2  
2 



Now, the displacements vectors A and B are added graphically using the proper scale and coordinate
system, and the graph is shown below:

The resultant vector is obtained by connecting the tail of the 1st vector and the head of the 2nd vector
and is formed as below:

The magnitude and direction of the resultant vector R = (14.7487, 42.0692) can be calculated
as follow:

R  Rx 2  Ry 2 

14.7487    42.0692 
2

2

 45

 Ry 
1  42.0692 
o
  tan 
  70
 14.7487 
 Rx 

  tan 1 

Finally, the % error between the graphical and component method is calculated by
For magnitude:

44.58  45
100%
44.58
 0.94%


For direction:



70.68o  70o

70.68o
 0.97%

 100%

Exercise 2:
It is given that the particle is exerted on by the following 3 forces:
F1=m1g @ 30o (m1 = 300 gr)
F2=m2g @ 110o (m2 = 450 gr)
F3=m1g @ 230o (m3 = 400 gr)
From above information, the force is calculated and the result can be tabulated as follow:
particle mass (kg) gravity(N) Force (N)
Direction (o)
1
0.3
2.94
2.94
30
2
0.45
4.41
4.41
110
3
0.4
3.92
3.92
230

Finding resultant force using Component Method
Each force F1, F2, and F3 in unit vector notation by taking the origin to be at the center of
the force table (at pivot point) with the +x axis along 0o and +y-axis along 90o, and the result is obtained
as below:
 3
1
F1  F1x x  F1 y y = 2.94cos  30o  x  2.94sin  30o  y = 2.94 
 x  2.94   y = 2.55x  1.47y
2
 2 

F2  F2 x x  F2 y y = 4.41cos 110o  x  4.41sin 110o  y = 2.94  0.432  x  2.94  0.940  y =  1.51x  2.76y
F3  F3 x x  F3 y y = 3.92cos  230o  x  3.92sin  230o  y = 3.92  0.463 x  3.92  0.766  y =  1.81x  3.00y

Now, the resultant force vector Fcomp can be obtained by using the component method, and its
magnitude and direction are calculated as follow:

...


Anonymous
Just what I needed…Fantastic!

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