college physics and intefral calculus problems

Math Correlation

Physics 1 And Physics 2

PHYSICS 1 PHYSICS 2

Physics 1ExponentsConversion Of UnitsDistance And DisplacementsMethods Of Vector

(Component, Graphical)Parallelogram MethodSpeed And Velocity Next Page

HOME

Graphical Analysis Of Velocity And AccelerationFree Falling BodiesForce And Newton’s Law Of MotionNewton’s Second LawGravitation AttractionNewton’s Law Of MotionNewton’s Third LawForces In EquilibriumSecond Condition Of Equilibrium

HOME

Physics 2KinematicsElectrons And Electronic FieldTemperature ConversionResistanceCapacitanceLinear Thermal ExpansionVolume Thermal Expansion

HOME

Exponents

1. (2m2) (3m) = 6m3

2. (5n3)) (4n2) = 20n5

3. (4s3) (4s4) = 16s8

4. (6o3) (2o4) = 12o7

5. (4a2)(2a3) = 8a5

6. (8c4)(3c2) = 24c6

7. 25cm3 = 5cm

5cm2 8. 60cm5= 5cm3

12cm2

9. 36cm3 = 6cm 6cm2

10. (10km2 )2 = (5km) 2 = 25km2

2km2

Conversion Of Units

1.) C.F. 1 inch = 2.54cmConvert 15 inches in cm.

15 inches x 2.54 cm = 38.1 cm 1 inch

2.) C.F. 1 mile = 5280 feet 1 kilometer = 1000 meters 1 inch = 2.54 centimeters

1 foot = 12 inches 1 meter = 100 centimeters

Convert 2 miles in kilometers. 2 miles x 5280 feet x 12 inches x 2.54 cm x 1 meter x 1 km 1 mile 1 foot 1 inch 100 cm 1000m

= 3.22 kilometers

3.) mass 560 milligrams = 0.02 ounce

560 milligrams x 1 gram x 1 ounce 1000 mg 28.350 grams

= 0.02ounce

4.) 73 km meters 73km x 1000meters = 73000 meters

1km

5.) 879 meters yards 879meters x 1km x 1mile x 1760yards

1000m 1.61km 1 mile = 960.9 yards6.)860 miles feet

860miles x 5280feet = = 4,540,800 feet 1mile

7.) 11 grams pounds 11 grams x 1pound = 0.02 pound

453.593grams

8.)266 liters gallons266 liters x 0.264gallons = 70.22gallons

1 liter9.)263 quarts fluid ounces

263quarts x 1gal x 231 cu.in x 128 fluid ounces 4 quarts 1 gal 231 cu.in

= 8416 fluid ounces10.) 131 inches meters 131 inches x 1 foot x 0.3048meters = 3.33meters

12 in 1 foot

Distances and Diplacement

• Tina walked 5km eastward to maries house

a. 30mb.D = displacement

distance = 30m displacement = 30m east

• Supposed Tina walked again 2.5km north from Marie's house to the church.

• A physics teacher walks 4 meters east, 2 meters south, 4 meters west, and finally 2 meters north. Determine distance and displacement.

METHODS OF VECTOR (GRAPHICAL METHOD)

METHODS OF VECTOR (COMPONENT METHOD)

• Ronald plans to visit the dentist clinic, which 30m, 34° WS from him. How far W and how far S is the clinic from his place?

• A motorcycle drove 155km, 70° ES. Find x and y component.

• The three finalists in a contest are brought to the center of the large flat field. Each is given a meter stick, a compass, a calculator, a shovel, and (in a different order of each contestant) the following three displacements.

72.4m, 32deg east of north57.3m, 36deg south of west17.8m, straight south

The three displacements lead to the point where the

keys to a new Porsche are buried . two contestants start measuring immediately, but the winner first calculates where to go. What does he calculates?

• Given the three displacements. Solve for the resultant and the angle.

d1 = 900 N, 30° SE

d2 = 750 N, 40° NE

d3 = 500 N, 50° NW

Solution:

Summation

Parallelogram Method

• Given three displacements A = 60m, 32° NWB = 63m, 24° SEC = 45m, 36° SW

P = B – 24°P = 68.42° - 24°P = 44.42°

90 – 44.42 = 45.48°90-45.58° + 36° = 8.42°

Speed And Velocity

• How far does a jogger run in 1.5hrs. (5400s) if his average speed is 2.22m/s?

s = 2.22 m/st = 5,400 s

S = d / td = s t = 2.22 m/s (5400s)d = 11988 m

• The first car travels from left to right and cover a distance of 1609 m in time of 4.740 sec.

• In the reverse direction, the car covers the same distance in 4.695s. From these data determine the average velocity for each run.

Acceleration and Increasing Velocity

• Suppose the plane in figure below starts from rest (Vo = 0 km /h) when to = 0s. The plane accelerates down the runway, and at time t = 29s attains a velocity of V = +260 km/h; determine the average acceleration of the plane.

Given: 8.97km/h

a= Δv a=89.7km/h ΔtΔv= 9(t) = 8.97km/h (29s) to=0 km/s

s t1 =1s v= 9 km/hΔv= 260.13 km/h t2 =2s v= 18 km/h

t3 =3s v= 27 km/ha= Δv= V - Vo 9x29= 261km/h Δt t - to

261 km/h x 1000m x 1hr = 72.5 m/s

1km 3600s =72.5 m/s – o m/s

29s- 0s a= 25 m/s2

Acceleration and Decreasing Velocity

• A drag racer crosses the finish line, and the driver replays a parachute and applies the brakes to slow down. The driver begins slowing down when t1 = 9.0s and the cars velocity is V1= +2m m/s. when t = 12.0s, the velocity has been reduced to V = 13m/s. What is the average acceleration of the dragster?

a= Δv = V – Vo Δt t – to =13m/s – 28m/s

12s – 9s = -5 m/s2

Solution:

Average and Instantaneous Velocities.

• A cheetah is crouched in ambush 20m to the east of an observer’s blind. At time t = 0s the cheetah charges an antelope in a clearing 50m east of the observer. The cheetah runs along a straight line. Later analysis of a video tape shows that during the first 2.0 s of the attack, the cheetah’s coordinate x varies with time according to the equation x = 20m + (5.0 m/s2) t2.

a.) find the displacement of the cheetah during the interval between t1 = 1s and t2 = 2s.

Soln: t 1=1s X1 = 20m + (5.0 m/s2)(1.0s)2

X1 = 25m

t2 = 2s x2 = 20m (5.0 m/s2)(1.0s)2

x2 = 40m

Δx = x2 – x1

= 40m- 25mΔ x= 15m

Solution:

Average and Instantaneous Accelerations.

• Suppose the velocity V of the car at any given t is given but the equation V = 60m/s t (0.50m/s3)(1s)2 . Find the change in velocity of the car in the time interval between t1 = 1.0s and t2 = 3.0s

t1= 1s t2= 3s V1= 60m/s +(0.50 m/s3)(1s)2 V2= 60m/s + (0.50m/s3)(3s)2

V1 = 60.50 m/s V2 = 64.50 m/s

V = V2 – V1

= 64.5m/s – 60.5 m/s

V = 4m/s

Displacement Of A Speedboat• The speedboat has a constant acceleration of

2.0m/s2. If the initial velocity of the boat is 6.0m/s. Find its displacement after 8.0sec.Given: A= 2.0m/s2

Vo = 6.0 m/s2

T = 8.0sec. Req’d: x = ?

Soln : V = Vo + at

= 6.0 m/s + 2.0 m/s2 (8.0s) V = 22 m/s

X = ½ (Vo + V) t

= ½ (6.0 m/s + 22 m/s)(8.0s) X = 122m

Catapulating Jet• A jet is taking off from the deck of an aircraft carrier.

Starting from rest, the jet is catapulated with a constant acceleration of 31 m/s2 along a straight line and reaches velocity of +62 m/s. Find the displacement of the jet.Given: a = 31m/s2

V = 62 m/s V0 = 0 m/s

Reqd : x

Solution : V = Vo + at

t = V – Vo

at = 62 m/s – 0 m/s

31 m/s2

t = 2.0s

x = ½ (Vo+ V) t

= ½ (0 m/s + 62 m/s)(2.0s)x = 62m

An Accelerating Spacecraft• The spacecraft is traveling with a velocity of 3250

m/s. suddenly, the retrorockets are fired, and the spacecraft begins to slow down with acceleration where magnitude is 10 m/s². What is the velocity when the displacement of the craft is 215 km., relative to the point where the retrorockets began firing?Given: Vo= 3250 m/s

a = -10 m/s² x = 215 km (1000m) = 215000 m

1km Reqd: V = ?

Solution: V = (Vo² + 2ax) ½

V = [(3250 m/s) ² + 2 (-10m/s²) (215000m)] ½

V = 2502.5 m/s

A motorcycle ride• A motorcycle starting from rest has an acceleration

of 2.6 m/s². After the motorcycle has traveled a distance of 120 m, it slows down with an acceleration of -1.5 m/s² until its velocity is 12 m/s. What is the total displacement of the motorcycle?

Given: Vo = 0

A = 2.6 m/s² X = 120 m V = 12 m/s

Solution : V= [Vo + 2ax] ½

V= [2ax] ½

V= [2(2.6 m/s²)(120m)] ½

V= 24.98 m/sX= V² - Vo² / 2a

X= (12 m/s)² - (24.98 m/s)² 2(-1.5 m/s²)

X = 160m

Graphical Analysis Of Velocity And Acceleration

V = Δx = x2 – x1 Δt t2 – t1

= 12 – 4 3 – 1

V = 4 m/s

• Given the ratio below solve for the velocity.

A Bicycle Trip• A bicycle maintains a constant velocity on the

outgoing leg of journey, zero velocity while stopped for lunch and another constant velocity on the way back. Figure below shows the position – time graph for such a trip. Using the position – time graph intervals indicated in the drawing obtain the velocities for each segment of the trip. Vo

Solution:V1= Δx = 400m = 2m/s

Δt 200sV2= Δx = 0 m = 0 m/s

Δt = 400s V3= Δx = -400m = -1m/s Δt 400s

Free Falling Bodies• A stone is dropped from rest from the top of a

building after 3s of free fall: a) what is the displacement of the stone? b) After 3s of free fall, what is the velocity V of the stone?

Δy= -9.8m/s2

Vy= ? Voy’= 0m/s

y= ? t= 3.0s

Solution:

a) y = Vot + ½ at² (take note: Vo = 0)

= ½ (-9.8m/s²)(3.0s)² = -44.1m

b) V = Vo + at

= (-9.8m/s²)(3.0s) = -29.4m/s

• A football game customarily begins with a coin toss to determine who kicks off. The referee tosses the coin up with an initial speed of 6 m/s in the absence of air resistance.a) How high does the coin go above its points of release?b) What is the total time the coin is in the air before returning to its point?

dy=-9.8m/s2 y = ?Vy= ? t = ?

Voy’= 6 m/s

Solution:a) y = V² - Vo² / 2a

y = (0 m/s) ² - (6 m/s)² / 2(-9.8m/s²) y = 1.84m

b) t = Vy – Vo / d

t = (0 m/s) – (6m/s) / 9.8m/s² t = 0.61s t = 0.61 x 2 t = 1.22s

• For instance in the figure below of y = 1.5 m. assuming that the initial velocity is Vo = 12 m/s. Determine the velocity during upward and downward motion. V = 0 m/s, t =4s

Solution:Upward

Voy = 12m/s

Dy = -9.8 m/sY = 1.55 V = [Vo + 2ay]½

V = [(12m/s + 2(9.8)(1.55))]½V = 10.71 m/s

DownwardVoy = -12m/s

Dy = -9.8m/s² Y = -1.5s V = [(-12m/s) – 2(-9.8m/s²)(-1.55)]½ V = 10.71m/s

A moving spacecraft• In the x direction, the spacecraft in the figure below

has an initial velocity component of Vox = 22 m/s. and an acceleration component of Ax = 24 m/s². In the y direction, the angles quantities are Voy = 14 m/s and Ay = 12 m/s². After the y direction of 7.0 s, find?

a) x and Vx

b) y and Vy

a) x Ax Vox Vx x t

24m/s 22m/s ? ? ?

x = Vot + ½ at²

= (22m/s)(7.0s) + ½(24)(7.0s)²x = 742mVx = Vox + Axt

= 22 m/s + (24m/s²)(7.0s)Vx = 19m/s

b) Ay = 12m/s², Voy = 14m/s, t = 7.0s

y = Voy + ½ Ayt

y = (14m/s)(7.0s) + ½ (12m/s²)(7.0s) y = 98m/s

Force And Newton’s Law Of Motion

• Pushing a stalled car Two people are pushing a stalled car. The mass of the car is 1850 kg. One person applies a force of 275 N of the car, while the other force of 395 N. Both forces act in the same direction. A third force of 560 N also acts on the car, but in a direction opposite to that in which the people are pushing. This force arises because of friction and the extent to which the pavement opposes the motion of the tires. Find the acceleration of a car.

Solution:A = EF / M take note : N = kg m/s²EF = F1 + F2 + F3

= 275N + 395N + (-560N)EF = 110N

A = 110N / 1850kgA = 0.06 m/s²

Neglecting Friction• Determine the horizontal force needed to accelerate

a 31 kg wooden cart from rest to a velocity of .56 m/s in 1.5s.Given: M = 31kg Vf= 0.56m/s

Vo = 0m/s T = 1.5s

A = V-Vo F = MA t EF = 31kg(0.37m/s²)

A = 0.56m/s – 0m/s EF = 11.57N 1.5s

A = 0.37m/s²

Applying Newton’s Second Law of Motion

• A man is stranded on a raft (mass of man and raft) = 1300 kg. By padding, he causes an average force F of 17 N to be applied to the raft in a direction due east. The wind also exerts a force A on the raft. This force has a magnitude of 15 N and points 67o north of east.a.) Ignoring any resistance from the water, find the x and y components of the raft’s acceleration.

Solution:Ay: Ax:

Sin67° = Ay / Ax cos67° = Ax / A Ay = A sin 67° Ax = 15N cos67°

Ay = 15N sin67° Ax = 5.86NSummation:

EF X YP 17N 0 NA 5.86N 13.81NEF 22.86N 13.81N

Ax = EFx / m Ay = EFy / m = 22.86N / 1300k = 13.81 / 1300kg

Ax = 0.02m/s² Ay = 0.01m/s²

• Suppose that the mass of the spacecraft is Ms = 11000 kg and that the mass of the astronaut is Ma = 92 kg. In addition, assuming that the astronaut exerts a force of P = +36 N on the spacecraft. Find the accelerations of the spacecraft and the astronaut.Aa = F / Ma = Ps / M = -36N / 92kgAa = -0.39m/s²

As = Pa / MsAs = 36N / 11000kgAs = 0.00327m/s²

Gravitational Attraction• What is the magnitude of the gravitational force that acts

on each particle assuming M1 = 12 kg (approximately the mass of a bicycle) M2 = 25 kg and r = 1.2 m?

M1 = 12kg

M2 = 25kg

r = 1.2m

F = G M1 M2 / r²

= (6.67259 x 10-11 N m²/kg²) (12kg)(25kg)(1.2m)²

F = 1.39 x 10-8 N

The Hubble Space Telescope• The mass of the Hubble Space Telescope is 11,600

kg. Determine the weight of the telescope.a) When it was resting on the earthb) as it in its orbit 598 km above the earth’s surface.

Me = 5.98 x 1024 kgRe = 6.38 x 106 M

Solution:a) M = 11600kg W = Mg

G = G Me / re

W = G Me M / re

= (6.67259 x 10-11 N m²/kg²)[(5.98 x 1024kg)(11600kg)] 6.38 x 106 m W = 7.25 x 1011 N

b) W = G MeM / Re + r

W = (6.67259 x 10-11 N m²/kg²) [(5.98 x 1024 kg)(11600kg)] (6.38 x 106 m + 598000m)

W = 7.25 x 1011 N

Application Of Three Newton’s Law Of Motion.

• A fully loaded lockhead L-1011 with a mass of 2.17x105 kg accelerates at full throttle down a level runaway. The engines push with a combined constant horizontal net force of 753 KN. If the plane starts from rest, how far will it go during 33.5s that it takes to reach lift of f velocity?Given: M = 2.17 x 105 kg

EF = 753 Kn = 753000 Vo = 0 m/s

T = 33.5s

Solution:A = EF = 753000N (kg m/s²) m 2.17 x 105 kg

A = 3.47 m/s²

X = Vot + ½ at²X = ½ (3.47 m/s²)(33.5s)²X = 1947.10m

The Normal Force And Newton’s Third Law

• The normal force FN is one component of the forces that a surface exerts on an object with which it is contact, namely, the component that is perpendicular to its surface.

Fn =?

Fn = 50 N

• Given the figure below solve for Fn.

Fn = ?Fn = W + A = 15 N+ 11N = 26 N

• GIVEN: w = 80N suspended on two strings t1 and t2

both have an angle of 42°. Solve for t1 and t2.

Forces In Equlibrium

Solve for T1 and T2:

EFx = 0

EFy = 0

EFx = 0

T1x + T2x + Wx = 0

- T1 cos 42° + T2 cos 42° = 0

-0.74 + 0.74 - Equation 1EFy = 0

T1 sin 42° + T2 sin 42° - 80N = 0

0.67 + 0.67 = 80 - Equation 2

Eq’n 1 & 20.67 (-0.74 T1 + 0.74 T2 = 0)

0.74 ( 0.67 T1 + 0.67 T2 = 80)

-0.50 + 0.50 T2 =0

0.50 + 0.50 T2 = 59.2 Substitute t2 in eq’n 1

T2 = 59.2 -0.74 T1 + 0.74 T2 = 0

-0.74 T1 + 0.74(59.2) = 0

-0.74 T1 = -48.81

-0.74 - 0.74 T1 = 59.2

Second Condition of Equilibrium• A gymnast weighing 48N standing on a vault which

is 4m from one end of the vault exert to support the gymnast vector diagram.

• A car weighing 1500N on a bridge which is 12m from one of the bridge piers and 8m from the other. Determine the force each piers exert to support the car.

• A 60 kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410 respectively. That horizontal pushing force is required to :

a) Just before the crate start moving?b) Slide the crate across a dock at a constant speed?

Solution:Fn = W = mg

Fn = 60kg (9.8m/s2)

Fn = 588N

a) Fs = Ms (FN)

= 0.760(588N) Fs = 446.88N

b) Fk = MK(FN)

= 0.410 (588N) Fk = 241.08N

• 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.

Find: d = ??Given: a = +3.2 m/s2

t = 32.8 s vi = 0 m /s

Solution:d = v it + 0.5at2 d = (0 m / s)(32.8 s)+ 0.5(3.20 m/s2)(32.8 s)2

d = 1720 m

Kinematics

• A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car. Given: d = 110 m

t = 5.21 s vi = 0 m/s

Find: a = ?Solution:

d = vit + 0.5at2 110 m = (0 m / s)(5.21 s)+ 0.5(a)(5.21 s)2

110 m = (13.57 s2)a

a = (110 m)/(13.57 s2) a = 8.10 m/ s2

• Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall? Given: a = -9.8 m

t = 2.6 s vi = 0 m/s

Find: d = ?? Vf = ??

Soution:d = vit + 0.5 at2 d = (0 m / s)(2.6 s)+ 0.5(-9.8 m/s2)(2.6 s)2

d = 33 m

Vf = vi + at Vf = 0 + (-9.8 m/s2)(2.6 s) Vf = -25.5 m / s (- indicates direction)

• 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 = ?

Solution:a = (Delta v)/t a = (46.1 m/s - 18.5 m/s)/(2.47 s)a = 11.2 m/s2

d = vit + 0.5at2

d = (18.5 m/s)(2.47 s)+ 0.5(11.2 m/s2)(2.47 s)2

d = 45.7 m + 34.1 md = 79.8 m

(Note: the d can also be calculated using the equation vf2 = vi2 + 2ad)

• 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 = ??Solution:

d = vit + 0.5at2 -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

• Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance which the sled travels?

Given: vi = 0 m/s vf = 44 m/s t = 1.80 s

Find: a = ?? d = ??

Solution:a = (Delta v)/t a = (444 m/s - 0 m/s)/(1.80 s)a = 247 m/s2

d = vi(t) + 0.5(a)(t)2

d = (0 m/s)(1.80 s)+ 0.5(247 m/s2)(1.80 s)2

d = 0 m + 400 md = 400 m

(Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)

• An engineer is designing the runway for an airport. Of the planes which will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway? Given: vi = 0 m/s

vf = 65 m/s a = 3 m/s2

Find: d = ??Solution:

vf2 = vi2 + 2(a)(d) (65 m/s)2 = (0 m/s)2 + 2(3 m/s2)(d)4225 m2/s2 = (0 m/s)2 + (6 m/s2)(d)(4225 m2/s2)/(6 m/s2) = dd = 704 m

• How many electrons are there in one coulomb negative charge?

Solution:Q = 1C E = 1.60 x 10 -19 CN = q

e = 1C

1.60 x 10 -19 C

= 6.25 x 10 -18

Electron And Electronic Fields

• Two objects whose charges are 1.0 C and -1.0 C are separated by 1km. Find the magnitude of attractive force that other charge exerts on the other.

1C 1000m 1C Solution:

F12 = k q1 q2 r2

= 9 x 109 N. m2 /c2 . (1C)(1C) (1000)2

F12 = - 9000 N F21 = 9000 N

+ -

• Two point charges q1 = 25 n C and q = 75 n C separated by 3 m. Find the magnitude of the electric force that q1 exerted on q2.

• Solution: F12 = k q1 q2

r2

= 9 x 109 N. m2 /c2 . (25C)(75C) (3)2

F21 = - 1.88 uN

25c 3 m 75c

• Three charges of a line. Three point charges that lie along the x-axis in a vacuum. Determine the magnitude and the direction of the net electrostatic force on q1.

Solution: F12 = k q1 q2

r2

= 9 x 109 N. m2 /c2 . (4 uC)(3uC) (.20m)2

F12 = - 2.7 N F13 = 9 x 109 N. m2 /c2 . (7 uC)(3uC)

(.15m)2

F13 = 8.4 N

F = F12 + F13

F = -2.7 + 8.4 F = 5.7 N

• Three charges on a plane. Figure below shows three point charges that lie in x, y plane in a vacuum. Find the magnitude and direction of the net electrostatic force on q1.

Solution: F12 = k q1 q2

r2

= 9 x 109 N. m2 /c2 . (4 uC)(6uC) (0.15m)2

F12 = 9.6 N F13 = 9 x 109 N. m2 /c2 . (4 uC)(8uC)

(0.10m)2

F13 = 2.88 N

F12y = F12 sin 73° F12x = F12 cos 73° = 9.6 sin 73° = 9.6 cos 73°= 9.18 N = 2.81 N

F x-component y-component

F122.81N 9.18N

F1328.8N 0N

FN 31.61N 9.18N

F= √ x2 + y2

= √ (31.61)2 + (9.18)2

= 32.92 NӨ = tan-1 9.18

31.61Ө = 16.19 ° NE

• Three charges q1 = 1x10-9 C, q2 = -6x10-9 and q3= -8x10-9 are arranged in figure below. What is the net statement resultant force on q2 due to the other two charges?

• Solution:F31 = 9x109 N. m2/c2 (4x10-9) (8x10-9C)

(10m)2

= 2.88 nNF32 = 9x109 N.m2/c2 (8x10-9C) (6x10-9C)

(8m)2

= 6.75 nN

Component of F31

F31x = F31 cos 37 = 2.88 nN cos 37 = -2.30 nNF31y = F31 sin 37 = 2.88 sin 37 = 1.73 nN

Component of F12x

F12x = F12 cos 10

= 6.75 cos 10 = 6.65 nNF12y = F32 sin 10 = 6.75 sin 10 = 1.17 nN

Fx = 0

F31x + F32x = 0

2.30 + 6.656 = 0Fx = 4.35 nN

Fy = 0

F32y + F31y = 0

1.17 + 1.73 = 0Fy = 2.9nN

F = √ (Fx)2 + (Fy)2

= √ (4.35)2 + (2.9)2

= 5.23 nN

Ө = tan-1 2.9 nN 4.35 nN

Ө = 33.69 ° NE

• Two charges q1 = -8MC and q2 = 12MC are placed in the air. What is the resultant force on a third charge q3 = -4MC, placed on the third of the other two charges?

Solution: F32 = 9x109 N. m2/c2 (4 MC) (12 MC)

(6m)2

= -12000 MNF31 = 9x109 N.m2/c2 (4 MC) (8 MC)

(12m)2

= -12000 MN

F = F32 + F31

= -12000 MN + (-12000 MN)= -24000 MN

• The positive test charges is qo = 3x10-8 C and the expensed of a force F = 6x10-8 N and the directionshown in the fig.

a. Find the force per coloumb that the test charge expense.b. Using the result of part a, predict the force that a charge of 12x10-8

would expense if it replaced q0.

Solution:a. E = F/qo

= 6x10-8 N 3x10-8 = 2 N/C

b. F = E(qo)

= 2N/C (12x10-8 C) = 0.24MN

• There is an isolated point charge of q = 15MC in a vacuum using a test charge of qo = 0.80 MC determine the electric field at point which is 0.20m away.

Solution:F = K qq0

r2

= 9x109 N.m2/c2 (15MC) (0.80MC) (0.20m)2

= 2700000 MN

E = F/qo

= 2700000 0.80 MC = 3375000 M N/C

Temperature Conversion• Convert 40° to °C

°C = °F – 32 1.5 = 40 – 32 1.5 °C = 5.33 °C

• Convert 11 °C to °F

°F = 1.5 (°C) + 32°F = 1.5 (11) + 32°F = 48.5 °F

• Convert 20F to Kelvin

K = °F + 273K = 20 + 273K = 293 K

• Convert 25 °C to Kelvin

°F = 1.5 (°C) + 32°F = 1.5 (25) + 32°F = 69.5 °F

K = °F + 273K = 69.5 + 273K = 342.5 K

• Convert 403 K to °C

K = °F + 273°F = K – 273°F = 403 – 273°F = 130 °F

°C = °F – 32 1.5

°C = 130 – 32 1.5

°C = 65.33 °C

Resistance

• 2Ω

10V

Find IT

V = IRI = V/RI = 10V 2ΩI = 5 A

• 3Ω

20V 6Ω

9Ω

Find IT, RT

Solution: RT = 3 Ω + 6 Ω + 9 Ω

RT = 18 Ω

IT = V/R

IT = 20V

18 ΩIT = 1.11 A

• 12V 7Ω 15Ω

Find RT, IT

Solution:RT = 7 Ω (15 Ω)

7 Ω + 15 ΩRT = 4.77 Ω

IT = V/R = 12V 4.77 Ω = 2.51 A

5Ω

•

10Ω

When I = 16 mAFind VT

Solution:RT = 5Ω + 10ΩRT = 15Ω

V = IRV = 16 mA (15Ω)V = 0.24 V

•

24V 7Ω 19Ω

20Ω 17Ω 5Ω

Find RT, IT

Solution: R195 = 19Ω + 5Ω IT = V/R = 24Ω = 24V R19517 = 24Ω (17 Ω) 13.64Ω 24Ω + 17Ω IT = 1.76 A

= 9.95ΩR1951720 = 9.95Ω (20Ω) 9.95Ω + 20ΩR1951720 = 6.64ΩRT = 6.64Ω + 7ΩRT = 13.64Ω

Capacitance• Two spaced metal plates form a capacitor. The area of

each plate is 9. The spacing between the plates is 7. What is the capacitance C?

Solution: C = A s = 9

7

• A sheet of glass is put between a hot stone and a cooler one. The thickness of the glass is 8. The conductivity of the glass is 3. What is the thermal resistance R of the glass?

Solution: R = T

c = 8

3

• A rod travelling near the speed of light contracts. The rod's proper length is 5.The Lorentz factor is 9. What is the rod's length L' at that speed?

Solution: L' = L

f = 5

9

• A block is fastened to a spring attached to a wall. The spring constant is 4. The distance from equilibrium is 7. What is the force F exerted by the spring on the block?

Solution: F = - c(d)

= - 4(7)= - 28

• A coiled wire forms an inductor. The flux through the inductor is 4. The number of windings is 9. The current through the wire is 2. What is the inductance L of the inductor?

Solution: L = N f c = 9 . 4 2 = 18

Linear Thermal Expansion• Find the linear expansion of a 12m concrete material from

12°C to 25°C.Solution:

L = α Lo T

= 12 x 10-6 (12m)(25°C- 12°C ) = 1.872 x 10-3

• Find the new length of a material with an original length of 250 m. After its temperature changes from 16°C to 70°C

Solution:L = α Lo T

= 12 x 10-6 (12m)(25°C- 12°C ) = 1.872 x 10-3

• In a sunny morning, the temperature rises from 25°C to 31°C. find the thermal expansion of the 10km concrete pavement.

L = α Lo T

= 12x10-6 (10km)(31°C - 25°C)L = 0.72

• Having a thermal expansion of 0.96. find the original length of the concrete rail after subdue to a temperature 76°C to 106°C

L = α Lo T0.96L = (12x10-6(106°C - 76°C)

L = 2666.67

Volume Thermal Expansion

• Find the volume thermal expansion of a 16m3 mercury. After its temperature changes from 30°c to 60°c.

∆V= βV (∆T)

∆V= 18x10-5°c-1 (16m3)(60°c-30°c)∆V= 0.0864m3

• Find the new volume of 28m3 mercury after its temperature changes from 20°c to 45°c.

∆V= βV (∆T)∆V= 18x10-5°c-1 (28m3)(45°c-20°c)

∆V= 0.126m3

V= Vo + ∆V

= 28m3 + 0.126m3

V= 28.126m3