Physics

*UNIT I

· CHAPTER 1

CONVERSION OF UNITS

Conversion of units refers to conversion factors between different units of measurement for the same quantity.

1. If the density of an object is 2.87 x 10−4 lbs/cubic inch, what is its density in g/mL?

ANSWER :

` x g/m = (2.87 * 10-4 lb / 1 in 3 ) (454 g / 1 lb ) (1 in 3 / 2.54 3 cm 3 ) (1 cm 3 / 1 m l ) = 7.59 * 10 -3 g/ml

2. A box measures 3.12 ft in length, 0.0455 yd in width and 7.87 inches in height. What is its volume in cubic centimeters?

ANSWER:

the volume of a box is calculated thus; V = L x W x H.

First you have to convert all the dimensions to the same unit such as inches.

x in = 3.12 ft. (12 in / 1 ft. ) = 3.74 in

x in = 0.0455 yd ( 3 ft / 1 yd 0 ) ( 12 in / 1 ft ) = 1.64 in

V = 37.4 in x 1.64 in x 7.87 in = 483 in3

Note the question is asking for cm3. We know the conversion from in to cm. We can easily convert in3 to cm3 thus:

( 2.54 in / 1 ft ) 3 = ( 2.54 3 cm 3 / 1 in 3)

Thus, we can convert 483 in3 into cm3 as follows:

x cm 3 = 483 in 3 ( 2.54 3 cm 3 / 1 in 3 ) = 7.91 * 10 3 cm 3

3. Convert 3598 grams into pounds.

ANSWER:

x lb = 3598 g (1 lb / 454 g ) = 7.93 lb

· CHAPTER 2

SPEED,VELOCITY & ACCELERATION

Speed is the rate of change of distance with time.

As a scalar it has magnitude only.

Average speed …

is measured over a non-zero time interval and

is represented by the symbol v_ave or v (overline)

Instantaneous speed …

is the limit of average speed as the time interval approaches zero,

is the first derivative of distance with respect to time, and

is represented by the symbol v (italic)

Velocity is the rate of change of displacement with time.

As a vector it must be stated with both magnitude and direction.

Average velocity …

is measured over a non-zero time interval and

is represented by the symbol v_ave or v (overline)

Instantaneous velocity …

is the limit of average velocity as the time interval approaches zero,

is the first derivative of displacement with respect to time, and

is represented by the symbol v (boldface)

The various forms of speed and velocity are defined by the following equations …

v = Δs Δt		average speed		v = lim Δs = ds Δt → 0 Δt dt		instantaneous speed
v = Δr Δt		average velocity		v = lim Δr = dr Δt → 0 Δt dt		instantaneous velocity

The relation of speed to velocity

An object's average speed approaches the magnitude of its average velocity as the time interval approaches zero.

Δt → 0

⇒

v → |v|

The instantaneous speed of an object is the magnitude of its instantaneous velocity.

v = |v|

The SI unit of speed and velocity is the meter per second [m/s].

Acceleration

is a vector quantity which is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

The average acceleration (a) of any object over a given interval of time (t) can be calculated using the equation

1. Calculate the size of a light year.

Astronomical distances are sometimes so large that using meter as the unit is cumbersome. For really large distances the light year is used. A light year is the distance that light would travel in one year in a vacuum. Since the speed of light is fast, and a year is long, the light year is a pretty good unit for astronomy. One light year is about ten trillion meters as the following calculation shows.

v =	Δs
	Δt
Δs =	vΔt	= (3.0 × 108 m/s)(365.25 × 24 × 3600 s)
Δs =	9.46 × 1015 m

Since both the speed of light and the year have exact defined values in the International System of Units, the light year can be stated with an unnecessarily large number of significant digits.

v =	Δs
	Δt
Δs =	vΔt	= (299,792,458 m/s)(365.25 × 24 × 3600 s)
Δs =	9,460,730,472,580,800 m

2. I went for a walk one day. I walked north 6.0 km at 6.0 km/h and then west 10 km at 5.0 km/hr. Determine the average speed for the entire journey.

Δt =	Δs			v =	Δs
	v				Δt
Δt1 =	6.0 km		= 1.0 h	v =	6.0 km + 10 km		=	16 km
	6.0 km/h				1.0 h + 2.0 h			3.0 h
Δt2 =	10 km		= 2.0 h	v =	5.3 km/h
	5.0 km/h

3. A football coach paces back and forth along the sidelines. The diagram below shows several of coach's positions at various times. At each marked position, the coach makes a "U-turn" and moves in the opposite direction

Seymour has an average speed of (95 yd) / (10 min) = 9.5 yd/min

and an average velocity of (55 yd, left) / (10 min) = 5.5 yd/min, left

· CHAPTER 3

FREE FALL

Free fall is motion with no acceleration other than that provided by gravity. Since this definition does not specify velocity, it also applies to objects initially moving upward. Although strictly the definition excludes motion of an object subjected to aerodynamic drag, in nontechnical usage falling through an atmosphere without deployed parachute is also referred to as free fall.

1. It takes 0.210s for a dropped wrench to travel past a poster that is 1.35 meters tall. How high above the top of the poster was the wrench released?

a = (v_m - v_t)/t

9.81 = (6.43 - v_t)/(0.105)

v_t = 5.40 m/s

2.A small spherical stone is thrown directly upward with an initial speed of 5.00m/s, from some unknown height above the ground. If the stone strikes the ground 2.50s after it is thrown; a) What was the height of the point from which the stone was thrown? b) What is the velocity of the stone as it strikes the ground? c) What is the maximum height that the stone reached?

First we draw a sketch of the stone and its flight path; Let's choose the upward direction as positive and the ground level as y = 0.

These values of the variables we will use for parts a and b of the problem. To find the answer to part c we will need to redefine where the final and initial positions are (more about that after we solve a and b).

Solving equation 4 for the initial position we get:

Now let's look at part c. Notice that we are asked about an intermediate position that the sphere would reach sometime between the initial and final positions as we have defined them. It will be easier to consider part c as a separate problem with a new diagram as shown:

3.How long does it take a ball to reach the ground 7.0 m below, if it is thrown straight up with an initial speed of 2.00 m/s?

d = v_it + (0.5)at²
(7.0 m) = (-2.00)t + (0.5)(9.8)t²

t = {1.42, -1.01}

Since t <>

t = 1.42 s

· CHAPTER 4

IMPULSE & MOMENTUM

• Newton’s original “quantity of motion”

• a conserved quantity

• a vector

• Definition: The linear momentum p of a particle is its mass times its velocity:

• p º mv

• Momentum is a vector, since velocity is a vector.

• Units: kg m/s (no special name).

The total momentum of a system of particles is the vector sum of the momenta of the individual particles:

p_total = p₁ + p₂ + ... = m₁v₁ + m₂v₂ +

1. An engine of the orbital maneuvering system (OMS) on a space shuttle exerts a force of 30000 N j for 4.00 s, exhausting a negligible mass of fuel relative to the 95,000 kg mass of the shuttle.
(a) What is the impulse of the force for this 4.00 s?
(b) What is the shuttle's change in momentum from this impulse?
(c) What is the shuttle's change in velocity from this impulse?
(d) Why can't we find the resulting change in kinetic energy of the shuttle?

a. Impulse = Ft = 30000(4) = 120000 N-s
b. change in momentum = impulse = 120000 N-s
c. j = m∆v
120000 = (95000)∆v
∆v = 1.26 m/s
d. We do not know the initial velocity.

2.What is the momentum of of an object with m=2.00 kg and v-40.0 m/s?

p = mv

p = (2.00)(40.0) = 80.0 kg*m/s

3. A 1000 kg car accidentally drops from a crane and crashes at 30 m/s to the ground below and comes to an abrupt halt. What impulse acts on the car when it crashes?

j = mΔv assuming constant mass

j = 1000(-30) = -30 000 kg-m /s

*UNIT II

· CHAPTER 4

GRAVITATIONAL FORCE

gravitational force - (physics) the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface; "the more remote the body the less the gravity"; "the gravitation between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them"; "gravitation cannot be held responsible for people falling in love"--Albert Einstein

1.Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of object 1 was doubled, and if the distance between the objects was tripled, then what would be the new force of attraction between the two objects?

Answer: F = 3.56 units

If the mass of one object is doubled. then the force of attraction will be doubled as well. But this affect is more than offset by the tripling of the separation distance. Tripling the distance would cause the force to be decreased by a factor of 9 (3²). The net affect on force is that it decreased by a factor of 2/9.

F = (16 units) • 2 / 9 = 3.56 units

2.As a star ages, it it believed to undergo a variety of changes. One of the last phases of a star's life is to gravitationally collapse into a black hole. What will happen to the orbit of the planets of the solar system if our star (the Sun shrinks into a black hole)? (And of course, this assumes that the planets are unaffected by prior stages of the Sun's evolving stages.)

Answer: No affect

The shrinking of the sun into a black hole would not influence the amount of force with which the sun attracted the Earth since neither the mass of the sun nor the distance between the Earth's and sun's centers would change.

3.Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was tripled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects?

Answer: F = 36 units

If each mass is increased by a factor of 3, then force will be increased by a factor of 9 (3*3). But this affect is partly offset by the doubling of the distance. Doubling the distance would cause the force to be decreased by a factor of 4 (2²). the net affect on force is that it increased by 9/4.

F = (16 units) * 9 / 4 = 36 units

· CHAPTER 2

STRESS AND STRAIN

· Stress Terms

Stress is defined as force per unit area. It has the same units as pressure, and in fact pressure is one special variety of stress. However, stress is a much more complex quantity than pressure because it varies both with direction and with the surface it acts on.

· Compression

Stress that acts to shorten an object.

· Tension

Stress that acts to lengthen an object.

· Normal Stress

Stress that acts perpendicular to a surface. Can be either compressional or tensional.

· Shear

Stress that acts parallel to a surface. It can cause one object to slide over another. It also tends to deform originally rectangular objects into parallelograms. The most general definition is that shear acts to change the angles in an object.

· Hydrostatic

Stress (usually compressional) that is uniform in all directions. A scuba diver experiences hydrostatic stress. Stress in the earth is nearly hydrostatic. The term for uniform stress in the earth is lithostatic.

· Directed Stress

Stress that varies with direction. Stress under a stone slab is directed; there is a force in one direction but no counteracting forces perpendicular to it. This is why a person under a thick slab gets squashed but a scuba diver under the same pressure doesn't. The scuba diver feels the same force in all directions.

In geology we never see stress. We only see the results of stress as it deforms materials. Even if we were to use a strain gauge to measure in-situ stress in the rocks, we would not measure the stress itself. We would measure the deformation of the strain gauge (that's why it's called a "strain gauge") and use that to infer the stress.

· Strain Terms

Strain is defined as the amount of deformation an object experiences compared to its original size and shape. For example, if a block 10 cm on a side is deformed so that it becomes 9 cm long, the strain is (10-9)/10 or 0.1 (sometimes expressed in percent, in this case 10 percent.) Note that strain is dimensionless.

· Longitudinal or Linear Strain

Strain that changes the length of a line without changing its direction. Can be either compressional or tensional.

· Compression

Longitudinal strain that shortens an object.

· Tension

Longitudinal strain that lengthens an object.

· Shear

Strain that changes the angles of an object. Shear causes lines to rotate.

· Infinitesimal Strain

Strain that is tiny, a few percent or less. Allows a number of useful mathematical simplifications and approximations.

· Finite Strain

Strain larger than a few percent. Requires a more complicated mathematical treatment than infinitesimal strain.

· Homogeneous Strain

Uniform strain. Straight lines in the original object remain straight. Parallel lines remain parallel. Circles deform to ellipses. Note that this definition rules out folding, since an originally straight layer has to remain straight.

· Inhomogeneous Strain

How real geology behaves. Deformation varies from place to place. Lines may bend and do not necessarily remain parallel.

1.50N are applied to a wire with a radius of 1mm. The wire was 0.7m long, but is now 0.75m long. What is the Young's Modulus for the material the wire is made of?

2.What is the strain on a Twix bar (original length 10cm) if it is now 12cm long?

3.Another wire has a tensile strength of 70MPa, and breaks under 100N of force. What is the cross-sectional area of the wire just before breaking?

4.10N of force are exerted on a wire with cross-sectional area 0.5mm². How much stress is being exerted on the wire?

0.5mm² = 0.5 x (10^-3)²m² = 0.5 x 10^-6m²

5.Glass, a brittle material, fractures at a strain of 0.004 and a stress of 240 MPa. Sketch the stress-strain graph for glass.

*UNIT III

· CHAPTER 1

WAVE SPEED

The speed of a wave is a property of the medium - changing the speed actually requires a change in the medium itself. If the medium does not change as a wave travels, the wave speed is constant.

1.The musical note A above middle C has a frequency of 440 Hz. If the speed of sound is

known to be 350 m/s, what is the wavelength of this note?

Given: frequency, f = 440 Hz

wave speed, v = 350 m/s

Unknown: wavelength, λ = ? m

Formula

v = f × λ

λ = v/f

answer

λ = 350 m / s/440 Hz

λ = 0.80 m

2.A buoy bobs up and down in the ocean. The waves have a wavelength of 2.5 m, and they

pass the buoy at a speed of 4.0 m/s. What is the frequency of the waves? How much time

does it take for one wave to pass under the buoy?

Given: wavelength, λ = 2.5 m

wave speed, v = 4.0 m/s

Unknown: frequency, f = ? Hz

period, T = ? s

formula

v = f × λ

f =v/λ

T= 1/f

Answer

f = 4 0 . m /s/2.5 m

f = 1.6 Hz

T = 1/1 6 . Hz

T = 0.62 s

3.Waves in a lake are 6 m apart and pass a person on a raft every 2 s. What is the speed of the waves?

Given: wavelength, λ = 6 m

period, T = 2 s

Unknown: wave speed, v = ? m/s

Formula

f= 1/T

v = f × λ

answer

f= ½ s=0.5Hz

v = (0.5 Hz) × (6 m)

v = 3 m/s

· CHAPTER 2

DOPPLER EFFECT

A change in the observed frequency of a wave, as of sound or light, occurring when the source and observer are in motion relative to each other, with the frequency increasing when the source and observer approach each other and decreasing when they move apart. The motion of the source causes a real shift in frequency of the wave, while the motion of the observer produces only an apparent shift in frequency. Also called Doppler shift.

formula,

· Δλ is the apparent change in wavelength,

· λ is the original wavelength when the source is stationary,

· v is the velocity of the source with respect to the stationary observer, and

· v_w is the speed of the wave.

In this case, four steps are needed to calculate the apparent frequency.

1. Determine the original wavelength using

λ = v_w / f

2. Determine the apparent change in wavelength using

Δλ = λ * (v / v_w)

3. Determine the new apparent wavelength using

λ' = λ ± Δλ

preceding source:
wavelengths drawn apart
λ' = λ + Δλ

approaching source:
wavelengths crowded together
λ' = λ – Δλ

4. Determine the new apparent frequency using

f ' = v_w / λ'

· CHAPTER 3

MIRROR EQUATION

Concave Mirrors

Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a concave mirror. The use of these diagrams were demonstrated earlier in Lesson 3. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and object size. To obtain this type of numerical information, it is necessary to use the Mirror Equation and the Magnification Equation. The mirror equation expresses the quantitative relationship between the object distance (d_o), the image distance (d_i), and the focal length (f).

The equation is stated as follows:

1.A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size.

1/f = 1/do + 1/d_i

1/(15.2 cm) = 1/(45.7 cm) + 1/d_i

0.0658 cm^-1 = 0.0219 cm^-1 + 1/d_i

0.0439 cm^-1 = 1/d_i

d_i = 22.8 cm

2.A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. (NOTE: this is the same object and the same mirror, only this time the object is placed closer to the mirror.) Determine the image distance and the image size.

1/f = 1/do + 1/d_i

1/(15.2 cm) = 1/(8.3 cm) + 1/d_i

0.0658 cm^-1 = 0.120 cm^-1 + 1/d_i

-0.0547 cm^-1 = 1/d_i

d_i = -18.3 cm

3.Determine the image distance and image height for a 5.00-cm tall object placed 30.0 cm from a concave mirror having a focal length of 15.0 cm.

Answer: d_i = 30.0 cm and h_i = -5.0 cm

Use 1 / f = 1 / d_o + 1 / d_i where f =15 cm and d_o = 45 cm

Then use h_i / h_o = - d_i / d_o where h_o = 5 cm, d_o = 45 cm, and d_i = 30.0 cm

NOTE: this is "Case 2"

Convex Mirrors

Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a mirror. The use of these diagrams were demonstrated earlier in Lesson 3 and in Lesson 4. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and image size. To obtain this type of numerical information, it is necessary to use the Mirror Equation and the Magnification Equation. The mirror equation expresses the quantitative relationship between the object distance (d_o), the image distance (d_i), and the focal length (f). The equation is stated as follows:

1. A 4.0-cm tall light bulb is placed a distance of 35.5 cm from a convex mirror having a focal length of -12.2 cm. Determine the image distance and the image size.

1/f = 1/do + 1/d_i

1/(-12.2 cm) = 1/(35.5 cm) + 1/d_i

-0.0820 cm^-1 = 0.0282 cm^-1 + 1/d_i

0.110 cm^-1 = 1/d_i

- d_i = -9.08 cm

2. Determine the focal length of a convex mirror which produces an image which is 16.0 cm behind the mirror when the object is 28.5 cm from the mirror.

Answer: f = -36.6 cm

Use the equation 1 / f = 1 / d_o + 1 / d_i where d_o = 28.5 cm and d_i = -16.0 cm

3.A focal point is located 20.0 cm from a convex mirror. An object is placed 12 cm from the mirror. Determine the image distance

Answer: d_i = -7.5 cm

Use the equation 1 / f = 1 / d_o + 1 / d_i where f = - 20.0 cm and d_o = +12.0 cm