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SET MY PROBLEMS UP LIKE THIS THE ANSWERS ARE ALREADY THERE
Problems should clearly identify (label) the following (note the points assigned each):
Variables:
Relationships:
Comuputation:
ANSWER:
EXAMPLE:
How it looks when presented in reply to a problem:
VARIABLES:v (velocity) =25 m/sa (acceleration) = ? unknownt (time) = 10 sRELATIONSHIPS:a = v/t,where, a = acceleration in m/s^2, v=velocity in m/s, and t= time in seconds (s).COMPUTATIONS:a = v/ta = (25m/s)/10s = 2.5 m/s^2ANSWER: 2.5 m/s^2
Ray Williams
April 20, 2013
TRIDENT UNIVERSITY
Module 5 – Case
Heat and Thermodynamics
RAY WILLIAMS
4/24/2013
Module 5 – Case
Heat and Thermodynamics
1. How much energy (in kcal ) is required to raise the temperature of 1 L of water from 20 to 100C, and then turn it to steam at 100C?
VARIABLES:
q(joules)=1000grams
1 liter of water is 1000 milliliters
1L would weigh 1000 grams( ~ 36 oz )
q(joules) = mass * specific heat * (Temp. final – Temp. initial)
q = 1000 grams * 4.180 J/gC * (100 C – 20 C)
q = 334400 joules
And 1 cal = 4.186 J
So q in kcal, 79885.33 cal = 79.885 kcal
ANSWER: 79.885 kcal
2. A 2 kg rod of aluminum (c = 0.9 kJ/kg-C) at 90C is dropped into 10 L of water at 10C. What is the final temperature of the mixture?
Heat lost by the rod = heat gained by the water
q_lost = q_gain
mass_aluminium * sp.heat_aluminium *(Tf – Ti) = mass_water * sp.heat_water*(Tf-Ti)
1L = 1000 milliliter
1L = 1000g,
so 10L = 10000g
2kg* 0.9kJ/kgC * (90C – x) = 10000g * 4.180 J/gC * (x – 10C)
2000g * 0.9 J/gC * (90 – x) = 10000g * 4.180 J/gC * (x – 10C)
1800 (90 –x) = 41800 ( x – 10)
90 –x = 23.222 (x – 10)
90 –x = 23.222x – 232.22
23.222x + x = 232.222 + 90
24.222x = 322.222
X = 13.308 C
ANSWER: 13.308C
3. A 0.2 kg block of an unknown metal at 50C is immersed in 1L of water at 4C. The equilibrium temperature of the mixture is 35C. What is the specific heat of the metal, in kcal/kg-C ?
Heat lost by the rod = heat gained by the water
q_lost = q_gain
mass_metal * sp.heat_metal *(Tf – Ti) = mass_water * sp.heat_water*(Tf-Ti)
1L = 1000 milliliter
1L = 1000g,
0.2kg* x * (50C – 35C) = 1000g * 4.180 J/gC * (35C – 4C)
200g * x* (15C) = 1000g * 4.180 J/gC * (31C)
3000 gC * x = 129580 J
x = 129580 J / 3000 gC
X = 43.1933J/gC
ANSWER: 43.1933 kJ/kgC
4. A liter of gas, initially at a pressure of 500Pa, is compressed from 1.00 L to 0.25 L. During the compression process, heat is dissipated to maintain a constant temperature. What is the final pressure?
Pi = 500 Pa
Vi = 1L
Vf = 0.25L
PV = nRT
PiVi = nRT = 500Pa * 1L
Pf*Vf = nRT = Pf * 0.25L
Equating both,
500Pa * 1L = Pf * 0.25L
Pf = 2000Pa
Final pressure, Pf = 2000 Pa = 2 kPa
ANSWER: 2 kPa
5. A radioisotope thermoelectric generator (RTG) generates electricity by absorbing heat from a core containing a radioactive material (plutonium 238 is typical) and discharging it into the environment. An RTG installed on an unmanned sonobuoy, tethered to the ocean floor, has a core temperature of 150 C, and discharges heat into sea water at 10 C. What is its theoretical efficiency?
T_cold = 10C
T_hot = 150C
Theoretical efficiency = 1 – Tcold / Thot
Theoretical effieciency = 1 – (10/150) = 1-0.0667 = 93.333%
ANSWER: 93.333%
Ray Williams
April 13, 2013
TRIDENT UNIVERSITY
Module 4 – Case
Fluid Mechanics
Ray Williams
4/23/2013
Module 4 – Case
Fluid Mechanics
1. A laboratory receives a sample of a metallic alloy composed of aluminum and magnesium. The sample is a solid cylinder with a diameter of 2 cm and a length of 5 cm. Its mass is 15.70 g. What is its density? What is its specific gravity?
d = 2cm, r = 1cm, h = 5 cm
M = 15.70 g
Volume of alloy = πr2h = 15.707 cm3
Density_alloy = mass / volume = 15.70g / 15.707 cm3 = 1 g / cm3 = 1000 kg/m3
Specific gravity = Density_alloy / Density (of magnesium) = 1000kg/m3 / 1738 kg/m3 = 0.5754
ANSWER1: 1000kg/m3
ANSWER2: 0.5754
2. A boat has a dry weight of 3500 lbs. How many liters of water does it displace?
SHOW WORK: 3500lbs=1587.5734Kg
displace water =weight of boat=1587.5734 Kg=1587.5734 *1.03 L =1635.2 liters
ANSWER: 1635.2 liters
3. A steel pipe runs vertically from the floor of the ocean to the surface. The pipe is empty — that is, contains air at atmospheric pressure . What is the water pressure on the pipe 200 m below the surface? At that point, what is the force on 1 cm2 of the pipe wall? (Density of seawater = 1.03 kg/L)
SHOW WORK: P @200 m below=P0+density*g*h =1.01325*10^5 +(1030*9.81*200) =2122185 N/m2 =20.944 atm force on pipe wall=P*A =2122185*(10^-4) =212.22 N
ANSWER1: 20.944 atm
ANSWER2: 212.22 N
4. Water is flowing through a circular pipe with an internal diameter of 1 meter, at a rate of 0.5 m3/s. What is the water’s flow velocity? What is its dynamic pressure?
SHOW WORK: water flow velocity=(0.5 m3/sec)/(pi/4)(1^2) v =0.6367 m/sec dynamic pressure=(1/2)*density*(v^2) =327.9 Kg/(m^2 sec)
ANSWER1: 0.6367 m/sec
ANSWER2: 327.9 Kg/(m^2 sec)
5. The static water pressure at point 1 in a pipe is 100 kPa, and the fluid velocity is 2 m/s. At point 2 downstream, the pipe widens and the velocity decreases to 1 m/s. What is the static pressure at that point? The usual assumptions concerning laminar flow, no friction losses, incompressibility, etc. all apply
SHOW WORK: P1+(1/2)*density*v1^2+density*gh=P2+(1/2)*density*v2^2+density*gh
* P1+(1/2)*density*v1^2=P2+(1/2)*density*v2^2
*100000+(1/2)*1030*(2^2)=P2+(1/2)*1030*(1^2)
* P2=101545 Pa
*P2=101.545 Pa
ANSWER: P2=101.545 Pa
Ray Williams
April 11, 2013
Module 2 – Case
Newton’s Laws
1. State and explain Newton’s Three Laws of motion, giving examples from daily life.
Answer: Law I). Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
EX. If a sled is going down a hill it will stay in motion until it hits the end of the hill or the pavement and then it stops
Law II)The relationship between an object’s mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.
EX. Is the amount of force we put on the bowling ball and the mass of the bowling ball determines its acceleration.
Law III)For every action there is an equal and opposite reaction.
EX. Is when we exercise and do pushups on the floor.
2. A 1500 kg car is accelerating at a constant rate. In two seconds, its velocity goes from 10
to 30 km/hr. What is the total force acting on the car, in N?
Answer: acceleration=(30-10)*10^3/(3600*2)=25/9 m/s^2
F=ma=1500*25/9=4166.7 N.
the total force on the car is around 4166.7 N.
3. An astronaut in his space suit has a total mass of 300 kg. His weight on the asteroid Ceres is 2.00 N. What is the acceleration of gravity on Erewhon?
Answer: a=F/m=2/300=6.67*10^(-3) m/s^2
therefore, the accelerate on of gravity on Erewhon is around 6.67*10^(-3) m/s^2
4. A 50 kg cannon shell is subjected to a constant acceleration of 1200 g while in the cannon barrel. What force is the burning propellant exerting on the shell?
Answer: F=ma=50*1200g=60000g N.
The force is around 60000g N.
5. Four forces, represented by four vectors, are plotted in standard position on the Cartesian plane. The lengths of the vectors, in arbitrary units, represent the strength of the forces in N. The forces are being applied to an object with a mass of 1 kg, located at the origin. In what direction will the object accelerate, and how rapidly? The force vectors are A(5,5) B(-3,-2) C(-1, 6) and D(-6, 3).
Answer: Fx=5-3-1-6=-5, Fy=5-2+6+3=12.
F=sqrt(5^2+12^2)=13 N..
acceleration=F/m=13/1=13 m/s^2
direction=arccos(-5/13)=112.62 degree.
the acceleration is 13 m/s^2 and the direction is 112.62 degree
Ray Williams
April 10, 2013
TRIDENT UNIVERSITY
Module 1 – Case
Displacement, Velocity, and Acceleration
Ray Williams
4/23/2013
Module 1 – Case
Displacement, Velocity, and Acceleration
1. A car is driving along a straight section of highway between Oklahoma City and Tulsa. At one point in time, the car is 50 miles from Tulsa. At a later point in time, it’s 40 miles from Tulsa. What is its displacement?
SHOW WORK :
Displacement is the change in vector position, that is to say, it’s simply the distance traveled, with a direction attached to it (a vector).
At point A, it was 50 miles from Tulsa, and at point B, it was 40 miles from Tulsa. This means it travelled 50-40=10 miles in total.
If we take point B, Tulsa, as our reference point, we can say that the car’s displacement was 10 miles [towards Tulsa]. Alternatively, if point A, Oklahoma, was our reference point, it would be 10 miles [away from Oklahoma].
ANSWER:10 miles
2. An ant is placed on a piece of graph paper. It crawls 3 cm on an azimuth (course angle) of 30 degrees, 5 cm on an azimuth of 180 degrees, and 2 cm on an azimuth of 270 degrees. At that point, the observer loses patience and swats and ant. What is the X-Y position of the squashed ant, measured with respect of the place where it started?
SHOW WORK:
Initially it went 3cm, 30 degrees. This means the horizontal distance he travelled, indicated by the light blue line in the diagram, must be 3cos30 (since cos 30 = adjacent / hypotenuse). In the diagram we also see that the ant moves to the left 2cm, horizontally, at some point in time. To find the total horizontal displacement therefore, we subtract the rightward horizontal displacement by the leftward horizontal displacement:
3cos30 – 2 = 2.598 – 2 = 0.598cm
This means the ant is 0.598cm right of its original position. This is x in the diagram.
To figure out y, the vertical distance the ant has travelled, we do something similar. We need to find the difference between the downward displacement (5cm) and the upward displacement (3sin30 this time):
3sin30 – 5 = 1.5 – 5 = -3.5cm
The negative, in this case, denotes that the ant is below the original starting point. As a result, y = 3.5cm
Therefore, the final position is [0.598, 3.5].
ANSWER: 0.598, 3.5
3. A motorcycle rider is attempting to set a speed record on a straight course. The cycle passes two points, a measured 100 m apart, in 1.2 seconds. What is its average velocity between those points?
SHOW WORK: Velocity is the displacement over time. 100m / 1.2 = 83.3m/s [forward].
ANSWER: 83.3m/s [forward].
4. A M1911 pistol fires a .45 ACP round. The bullet leaves the 13 cm barrel with a velocity of 300 m/s.
Answer: the bullet is not moving at the beginning of the 13cm barrel, then leaves the barrel at 300m/s. The average velocity would be the average of all velocities while it was inside the length of the barrel, and we can take two points for this measurement: beginning of the barrel where velocity was zero, and end of barrel which was 300m/s.
Avg. velocity = (300m – 0) / 0.13m = 2310m/s
a. The time the bullet takes to travel through the 13cm barrel, exiting at 300m/s while starting at 0m/s, requires calculation using a classic kinematic equation: d = (v1 + v2) /2 * t
b. where v1 = initial velocity = 0 d = displacement = 13cm t = time 0.13m = (0 + 300m/s) / 2 * t t = 0.13 / 150 = 8.67×10^-4 seconds
c. Avg. acceleration: v2^2 = v1^2(t) + 2ad
300^2 = 0 + 2a(0.13) a = 300^2 / 0.26 = 3.46×10^5m/s
ANSWER1: 2310m/s
ANSWER2: -4 seconds
ANSWER3: 3.46×10^5m/s
5. A small powder charge propels a shoulder-launched antitank missile from its tube at 20 m/s. At 0.2 seconds after launch, the rocket motor ignites and accelerates the missile at a constant 20 g. After 2 seconds, the missile hits its target and detonates. How far away was the target? Assume a straight flight path. (Hint: Calculate two distances, one before and one after rocket motor ignition.)
Answer: distance traveled prior to rocket launch:
initial velocity = 20m/s
t = 0.2s Velocity = displacement / time
d = vt = 20*0.2 = 4m
distance after rocket ignition:
t = 2s
a = 20g
initial velocity = 20m/s
d = ?
d = v1(t) + 1/2 (a)(t^2)
d = 20(2) + 1/2 (20×9.81)(2^2)
d = 40 + 392.4 = 432.4m
*add the two distances together:
4 + 432.4 = 436.4m
ANSWER: 436.4m
