Unit 1 Review

Mathematical introduction to physics

1.     Express  a:     v=  square root of  2as

v² = 2as à v²/2s = a

2.     Express t:      v =  v₀  +  at

(v-v₀)/a = t

3.     Express  c:      E  =  mc²

E/m = c² à sqrt(E/m) = c

Solve the following (include units)         N = Kgm/s²    j = Nm   W=j/s

                                   

4.     2.5x10¹² m/s²  x   3.3x10¹³ s  = (8.25x10²⁵)m/s

5.     3.5x10^-4kg  x  2.5x10^-10m/s² = (8.75x10^-14)N

6.     6.5x10⁵m/s  :    3.3x10^-3s  = (1.97x10⁸)m/s²

7.     2x10⁵N    :     3x10¹⁵m/s² = (6.67x10^-11)Kg

8.     2.4x10⁶w  *  2.3x10³s  = (5.52x10⁹)j

Can these equations be correct?  Use dimensional analysis of units to prove your point!   d (m)     v (m/s)      t (s)     a (m/s2)

1.     d = v/t   à m = (m/s)/s à m = m/s² NOT VALID    

2.     d = vt + at²/2 à m = (m/s)(s) + (m/s²)(s²)/2 à m = m + m/2 VALID

3.     v = 2ad à (m/s) = 2(m/s²)(m) à (m/s) = (m²/s²) NOT VALID

                                                                                     

Solve (your answer should be stated in MKS units)

1.     5.3x10³km + 20m  = 5300000m + 20m = 5300020m                   

2.     0.5m/s + 2km/h  =  0.5m/s + 0.56m/s = 1.06m/s

3.     5.5x10³ g + 20 kg = 5.5kg + 20kg = 25.5kg                          

4.     42cm + 2m  = 0.42m + 2m = 2.42m

 

Find the angles         a                               a= 15         b= 22

                                               b

∠A = tan^-1(15/22) = 34.3⁰                                              

c = 26.6

∠B = cos^-1(15/26.6) = 55.7⁰               b                                

                                                                         25⁰       

Find the sides a and  b          a                    

           25

a à sin(25⁰) = a/25 à a = 10.6

b à cos(25⁰) = b/25 à b = 22.7

Use units and dimensional analysis to find out if the formulas are valid

1.     Distance = speed * time à m = (m/s)*s à VALID                         

2.     acceleration = speed/time à (m/s²) = (m/s)/(s) à VALID

3.     Force = mass * acceleration à N = kg*(m/s²) à VALID                    

4.     distance = energy * time à m = (j)*(s) à NOT VALID

5.     Power = work * time à (j/s) = (j)*(s) à NOT VALID                              

6.     speed = power/force à (m/s) = (j/s)/(N) à VALID

The displacement of an object from a start point was checked at 2 s intervals. The results are in the following chart

Time (t) in (s)	Displacement (d) in (m)
0	0
2	4
4	8.5
6	12.2
8	15.8
10	21
12	25.3

a.     Graph the results using proper graphing techniques.

b.     What is the relationship you observe?

c.      Find the constant (slope) including units. What quantity does it represent?

d.     Write the equation of the line.

e.      Show one example of interpolation and one for extrapolation.

Practice Graphing

Here are my solutions to the three sets of data:

GRAPH 1:

This graph resembles an inverse relation between the x values (radius) and the y values (resistance). As the radius (mm) increases, the resistance (Ohms) decreases. Since this is a direct square relation, we cannot find a general slope for the graph. We can, however, use derivatives to find the slope at instantaneous points. To find the general slope of the curve (not accurate) we must first linearize the graph. Here is the same graph with a linear trend line:

To linearize this function I plotted 'y' vs '1/x' since this has an inverse relationship. As seen in the equation given, the slope of the line is m=(0.9667)Ohms/mm --> this means that the resistance increases by 0.9667Ohms for every 1 mm that the radius increases. The R² of the exponential trend line is 0.9254 while the R² of the linear trend line is 0.9984. This means the the linear trend line is much more accurate and better displays the expected results.

GRAPH 2:

This graph resembles an direct relation between the x values (Force) and the y values (Acceleration). As the Force (N) increases, the Acceleration (m/s²) increases in a linear manner. We use a trend line to find the equation of the graph. Using the equation we can find the slope, which is m=0.2902m/s²N = 0.2902 (m/s²)/(kgm/s²) = 0.2902kg-1--> the R² of the trend line is 0.9939, meaning that it is extremely accurate. The reason for this direct relation is because of the famous equation F=ma, where F is Force (N), m is mass (kg) and a is acceleration (m/s²). Given that the mass stays the same, as the force increases so must the acceleration to keep the equation balanced.

GRAPH 3:

This graph resembles an direct square relation between the x values (radius) and the y values (resistance). As the time (s) increases, the position (m) increases exponentially. Since this is a direct square relation, we cannot find a general slope for the graph. To find the general slope of the curve (not accurate) we must first linearize the graph. Here is the same graph with a linear trend line:

To linearize this function I plotted 'y' vs 'x^2'. As seen in the equation given, the slope of the line is m=(20.733)m/s² --> this means that the position increases by 20.733m for every 1 second that passes. The R² of the linear trend line is 0.8274 while the R² of the exponential trend line is 0.9372. This means the the exponential trend line is much more accurate and better displays the expected results.

Height of the Fence

We had an in-class "lab" today to try and measure the height of the fence using techniques available to scientists 500 years ago, such as Galileo. While standing 5m behind the fence, we used device called clinometer to measure the angle of elevation from our eyes to the top of the fence. Using the basic trigonometric function "tangent" we were able to calculate the height of the fence from the elevation of our eyes. We then measured the height of our eyes to the ground and added that to the final number. Daniel Neshto and I got the same results, 5.23, and we were the closest out of all the classes! No cheating involved.

Here are my calculations:

My height = 1.66m
Angle of elevation = 35.5 degrees 
Distance from fence = 5m

Calculations:
tan(35.5) = x/5
--> x = 5*tan(35.5) = 5.23

Variable Isolation - August 20, 2012

Isolating Variables

Michael Matias

August 20, 2012

I wasn't sure whether, when taking the square root of a number, the solution will be + and -. 

Example: v² = 2E/m --> v = sqrt(2E/m)

Are there two solutions to the equation, one with a negative value?

Solve the following equations from the variable(s) requested

1.       E = 1/2mv² for m and v

2E = mv² à 2E/m = v² à v = sqrt(2E/m)

2E = mv² à m = 2E/v²

2.       E = mgh for h

h = E/mg

3.       1/f = 1/d_o + 1/d_i for d_o

d_od_i = fd_i + fd_o à fd_i = d_od_i – fd_o à fd_i = d_o(d_i -f) à d_o = fd_i /(d_i -f)

4.       s_o/s_i = d_o/d_i for s_i

d_i s_o = d_o s_i à s_i = d_i s_o / d_o

5.       P=Fd/t for d

Pt = Fd à d = Pt/F

6.       E = hf -W_o for W_o and h

E – hf = - W_o à W_o = hf – E

E + W_o = hf à h = (E + W_o)/f

7.       v = square root of TL/m for T

v = sqrt(TL/m) à v² = TL/m à v²m = TL à T = v²m/L

8.       F = Kq₁q₂/r² for r

Fr² = Kq₁ q₂à r² = Kq₁ q₂/F à r = sqrt(K q₁ q₂/F)

9.       R = pl/A for A

AR = pl à A = pl/R

10.   a_c = mv²/r for v

a_c r = mv² à v² = a_c r/m à v = sqrt(a_c r/m)

Station Summaries - August 16th, 2012

Mechanics

Station 1

Summary: Two balls are thrown at the same time, one is let go without x-axis velocity but the second ball is thrown straight. Which ball will fall first?

Answer: Both will fall at the same time due to the constant acceleration due to gravity (9.8m/s²).

Station 2

Summary: Newtonian balls (study of momentum)

Station 3

Summary: A car has a propeller, which way will it move?

Answer: The opposite way where the air is blown. This is because of Newton’s third law of motion, for every action there is an equal and opposite reaction. As the propeller applies force on the air, the air applies the same force on the car which makes the car move the opposite way.

Station 4

Summary: Pendulum.

Question: What determines the amount of times it takes to make a complete swing?

Answer:  gravity, length of the rope

Wrong answers: force of pushing, mass of object

Summary: One of the pendulums starts swinging, will any of the others move as a result?

Answer: Yes. Due to similar natural frequencies, another pendulum with the same length will start moving as well. This is why opera singers can break glasses, because they manage to sing a pitch whose sounds waves match those of the glass.

Station 5

Summary: Balls roll in two different sloped structures. Which ball will fall faster?

Hypothesis: Same time because of their potential to reach the same max velocity.

Answer: The one with the steeper slope. The reason for this is that even though they both reach the same max velocity, that with the steeper slope reaches that velocity earlier.

Static Electricity

Station 6

Summary: A ruler is rubbed against another material to gain static electricity. After charged and when put next to water, the water moves toward the ruler.

Magnetism

Station 7

Summary:  2 big magnets are trying to touch each other where North touches North and South touches South.

Answer: Impossible. Opposite poles attract and similar poles repel.

Station 8

Summary: How do we make electricity?

Answer: All you need is a copper wire (or any other metal that is a good conductor of electricity) and a magnet. The three things that determine the amount of electricity produced is: speed of movement of magnet, size of magnet, amount of coil.

Trigonometry Review - August 16th, 2012

Trigonometry Review

Michael Matias

August 16, 2012

7) Solve for all sides and angles of the triangles:

Working With Units 3 - August 16th, 2012

Working with units

1.      Find out what is the radius (in m) and mass (in kg) of the earth. From the radius calculate the volume of the earth assuming it’s a perfect sphere. Then calculate the density of the earth in kg/m³.

radius of earth = 6378100m

mass of earth = 5.97219*10²⁴ kg

volume of earth = à V=4/3 πr^3 àV = 1.086*10²¹ m³

density of earth = 5499.254 kg/m³

2.    A car travels from Tel Aviv to Haifa a distance of 100km in 50 minutes. Assuming the car moves at a constant speed, calculate the speed in m/s. (Change distance to m and time to seconds)

distance = 100km = 100000m

rate = 50 minutes = 3000 seconds

speed = distance/rate = 33.33 m/s

3.    A student riding on a bike, changes its speed from 1m/s to 3m/s over a period of 10 seconds. What is the acceleration including units. (Acceleration means the change in speed per second)

Change in speed = 3m/s – 1m/s = 2m/s

Rate of change = 10s

Accelarion = change in speed / rate of change

Acceleration = (3-1)/10 = 2/10 = 0.2 m/s²

4.    A satellite circles the earth once every hour and a half. Find the speed of the satellite in m/s. (Assume the satellite revolves very close to the earth’s surface.)

Circumference of earth = 2πr = 2*3.14*6378100 = 40054468m

Speed = distance/time = 40054468m/90minutes = 445049.64m/minutes = 7417.49m/s

5.     A force of 200N accelerates a 20kg object from rest. What acceleration is it causing? (Show also how you have obtained the units for your answer) (Remember: Force = mass x acceleration.)

Force = mass*acceleration à 200N = 20kg*acceleration à acceleration = 10N/kg

N = kgm/s² à 10N/kg = 10m/s²

Working With Units/Confirming Equations - August 16th, 2012

Working with units/confirming equations

Quantity	Letter used in equations	unit
Length, distance	d	m (meter)
Time	t	s (second)
Mass	M	kg (kilogram)
velocity	v	m/s
acceleration	a	m/s2
Force	F	N (Newton)
Energy, work	W	J (joule)
power	P	w (watt)
Frequency	f	Hz

Using the above chart and dimensional analysis, determine if the following equations are valid:

d = vt + at²/2 à d(m) = v(m/s)*t(s) + a(m/s²)*t²(s²)/2 = m+m/2 VALID

W = vt à W(J) = v(m/s)*t(s) = m/s² NOT VALID

P = Wt àP(w) = W(J)*t(s) NOT VALID

P = Fv àP(w) = F(N)*v(m/s) VALID

W = Fd à W(J) = F(N)*d(m) VALID

v = at à v(m/s) = a(m/s²)*t(s) VALID

t = W/P à t(s) = W(J)/P(w) = W(Nm)/P(Nm/s) = s Valid

W = Pt à W(J) = P(w)*t(s) = P(J/s)*t(s) VALID

v² = 2ad à v²(m²/s²) = 2*a(m/s²)*d(m) VALID

W = Mv à W(J) = m(kg)*v(m/s) NOT VALID

a = vt à a(m/s²) = v(m/s)*t(s) NOT VALID

P = Mv/t à P(w) = m(kg)*v(m/s)/t(s) = kgm/s² NOT VALID

t = square root of (d/a) à t(s) = sqrt[d(m)/a(m/s²)] = s VALID

v² = W/M à v²(m²/s²) = W(J)/m(kg) VALID

F = Ma à F(N) = m(kg)*a(m/s²) VALID

P = Fd/t à P(w) = F(N)*d(m)/t(s) VALID

w= Mv²/2 à j/s = (kg)(m²/s²)/2 = 2kgm²/s² VALID

v= d*f à m/s = m*1/s = m/s VALID

f= a*t² à 1/s = m/s² * s² = m NOT VALID