OA02a: Thermal Physics and 1D Kinematics Due: 11:59pm on Wednesday, August 24, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy
Prelecture Concept Question 17.06
Part A If both the pressure and volume of a given sample of an ideal gas are doubled, what happens to the temperature of the gas in Kelvins? ANSWER: The temperature of the gas is reduced to one-half its original value. The temperature of the gas in increased by four times its original value. The temperature of the gas is reduced to one-fourth its original value. The temperature remains constant. The temperature of the gas is increased by two times its original value.
Correct
Prelecture Concept Question 17.04
Part A Which of the following statements are true? Check all that apply. ANSWER: Water contracts as it freezes at 0°C. Solid ice is less dense than liquid water. Liquid water expands with increasing temperature above 4°C. Liquid water expands with increasing temperature between 0°C and 4°C.
Correct
Calorimetry Conceptual Question A 0.5 block of aluminum ( insu sullate ted d tu tub b of co colld wate terr at 0 block blo ck are are mea measur sured ed to be 20 .
Part A
(
) is heated to t o 200 . The block is then quickly placed in an ) and se sea aled. At equilibrium, th the e te tem mperatu turre of th the e wate terr and
If the original original experiment is repeated repeated with a 1.0 block?
aluminum block, what is the final temperatur temperature e of of the water and
Hint 1. Heat energy and temperature The amount of heat heat energy energy needed neede d to change the temperature temperature of a substance depends depends on the mass of the substance , the specific heat of the substance , and the change in the temperatur temperature e of the substance. This is summarized mathematically as .
Hint 2. Conservation of energy The total amount of energy in the system must be conserved. Once the block is placed in contact with the cold water, heat energy will flow from the hot test block into the cold water, lowering the temperature of the block and raising the temperature of the water. A larger block can transfer a larger amount of heat energy for a given change in temperature. This will result in a larger equilibrium temperature.
ANSWER: less than 20 20 greater than 20
Correct
Part B If th the e original exp xpe erime men nt is repeate ted d with a 1.0 temperature of the water and block?
cop co pper (
) block ck,, what is th the e fi fin nal
ANSWER: less than 20 20 greater than 20
Correct
Part C If the or origin iginal al exper experimen imentt is rep repeate eated d but 100 temperature of the water and block?
of the 0
water wa ter is rep replaced laced with 100
of 0
ice, wh what at is the final
Hint 1. Latent heat of fusion The latent heat heat of of fusion is the amount of heat heat energy energy neede needed d to melt 1.0 fusion of ice is 334,000 .
ANSWER:
of a substance. The latent heat heat of of
less than 20 20 greater than 20
Correct
Part D If the original experiment is repeated but 100 temperature of the water and block?
of the 0
water is replaced with only 25
of 0
ice, what is the final
Hint 1. Find the equilibrium temperature In the original experiment, the temperature of 100 of 0 water was raised by 20 . The amount of heat required to do this can be found as follows. For specific heat and change in temperature , , or . Therefore,
.
Is more or less energy than this required to melt and heat 25
of ice to 20
?
Hint 1. How to approach the problem The amount of energy required to melt and heat 25 of ice to 20 can be broken into two parts: the energy required to melt the ice and the energy required to heat the water after the ice melts.
ANSWER: more less
ANSWER: less than 20 20 greater than 20
Correct
Problem 19.9
Part A
How much energy is required to bring a 1.6
pot of water at 24
to 100
?
Express your answer using two significant figures. ANSWER: = 5.1×105
Correct
Part B For how long could this amount of energy run a 100-
lightbulb?
Express your answer using two significant figures. ANSWER: = 85
Correct
Problem 19.20 A 60
ice cube at its melting point is dropped into an insulated container of liquid nitrogen.
Part A How much nitrogen evaporates if it is at its boiling point of 77 and has a latent heat of vaporization of 200 Assume for simplicit y that the specific heat of ice is a constant and is equal to its value near its melting point.
?
Express your answer using two significant figures. ANSWER: =
0.12
Correct
± Average Velocity from a Position vs. Time Graph Learning Goal: To learn to read a graph of position versus time and to calculate average velocity. In this problem you will determine the average velocity of a moving object from the graph of its position as a function of time . A traveling object might move at different speeds and in different directions during an interval of time, but if we ask at what constant velocity the object would have to travel to achieve the same displacement over the given time interval, that is what we call the object's average velocity . We will use the notation to indicate average velocity over the time interval from to . For instance, is the average velocity over the time interval from to .
Part A Consulting the graph shown in the figure, find the object's average velocity over the time interval from 0 to 1 second. Answer to the nearest integer.
Hint 1. Definition of average velocity Average v elocity is defined as the constant velocity at which an object would have to travel to achieve a given displacement (difference between final and initial positions, which can be negative) over a given time interval, from the initial time to the final time . The average velocity is therefore equal to the displacement divided by the given time interval. In symbolic form, average velocity is given by .
ANSWER: = 0
Correct
Part B Find the average velocity over the time interval from 1 to 3 seconds. Express your answer in meters per second to the nearest integer.
Hint 1. Find the change in position The final and initial positions can be read off the y axis of the graph. What is the displacement during the time interval from 1 to 3 seconds? Express your answer numerically, in meters ANSWER: = 40
Average v elocity is defined as the constant velocity at which an object would have to travel to achieve a given displacement (difference between final and initial positions, which can be negative) over a given time interval, from the initial time to the final time . The average velocity is therefore equal to the displacement divided by the given time interval. In symbolic form, average velocity is given by .
ANSWER: = 20
Correct A note about ins tantaneous velocity. The ins tantaneous velocity at a certain moment in time is represented by the slope of the graph at that moment. For straight-line graphs, the (instantaneous) velocity remains constant over the interval, so the instantaneous velocity at any time during an interval is the same as the average velocity over that interval. For instance, in this case, the instantaneous velocity at any time from 1 to 3 seconds is the same as the average velocity of .
Part C Now find
.
Give your answer to three significant figures.
Hint 1. A note on the displacement Since the object's position remains constant from time 0 to time 1, the object's displacement from 0 to 3 is the same as in Part B. However, the time interval has changed.
ANSWER: = 13.3
Correct Note that
is not equal to the simple arithmetic average of
and
, because they are averages for time intervals of different lengths.
Part D Find the average velocity over the time interval from 3 to 6 seconds. Express your answer to three significant figures.
Hint 1. Determine the displacement What is the displacement? Answer to the nearest integer. ANSWER:
, i.e.,
= -40
Hint 2. Determine the time interval What is the time interval? Answer to two significant figures. ANSWER: = 3.0
ANSWER: = -13.3
Correct
Part E Finally, find the average velocity over the whole time interval shown in the graph. Express your answer to three significant figures.
Hint 1. Determine the displacement What is the displacement? Answer to the nearest integer. ANSWER: = 0
ANSWER: = 0
Correct Note that though the average velocity is zero for this time interval, the instantaneous velocity (i.e., the slope of the graph) has several different values (positive, negative, zero) during this time interval. Note as well that since average velocity over a time interval is defined as the change in position (displacement) in the given interval divided by the time, the object can travel a great distance (here 80 meters) and still have zero average velocity, since it ended up exactly where it started. Therefore, zero average velocity does not necessarily mean that the object was standing still the entire time!
What Velocity vs. Time Graphs Can Tell You
A common graphical representation of motion along a straight line is the v vs. t graph, that is, the graph of (instantaneous) velocity as a function of time. In this graph, time is plotted on the horizontal axis and velocity on the vertical axis. Note that by definition, velocity and acceleration are vector quantities. In straight-line motion, however, these vectors have only a single nonzero component in the direction of motion. Thus, in this problem, we will call the velocity and the acceleration, even though they are really the components of the velocity and acceleration vectors in the direction of motion, respectively. Here is a plot of velocity versus time for a particle that travels along a straight line with a varying velocity. Refer to this plot to answer the following questions.
Part A What is the initial velocity of the particle,
?
Express your answer in meters per second.
Hint 1. Initial velocity The initial velocity is the velocity at
.
Hint 2. How to read a v vs. t graph Recall that in a graph of velocity versus time, time is plotted on the horizontal axis and velocity on the vertical axis. For example, in the plot shown in the figure, at .
ANSWER: = 0.5
Correct
Part B What is the total distance
traveled by the particle?
Express your answer in meters.
Hint 1. How to approach the problem Recall that the area of the region that extends over a time interval under the v vs. t curve is always equal to the distance traveled in . Thus, to calculate the total distance, you need to find the area of the entire region under the v vs. t curve. In the case at hand, the entire region under the v vs. t curve is not an elementary geometrical figure, but rather a combination of triangles and rectangles.
Hint 2. Find the distance traveled in the first 20.0 seconds What is the distance
traveled in the first 20 seconds of motion, between
and
?
Express your answer in meters.
Hint 1. Area of the region under the v vs. t curve The region under the v vs. t curve between and dimensions by , and a triangle of base figure.
can be divided into a rectangle of and height , as shown in the
ANSWER: = 25
Hint 3. Find the distance traveled in the second 20.0 seconds What is the distance
traveled in the second 20 seconds of motion, from
to
Express your answer in meters.
Hint 1. Area of the region under the v vs. t curve The region under the v vs. t curve between by , as shown in the figure.
and
is a rectangle of dimensions
?
ANSWER: = 40
Hint 4. Find the distance traveled in the last 10.0 seconds What is the distance
traveled in the last 10 seconds of motion, from
to
?
Express your answer in meters.
Hint 1. Area of the region under the v vs. t curve The region under the v vs. t curve between height , as shown in the figure.
and
is a triangle of base
and
ANSWER: = 10
ANSWER: = 75
Correct
Part C What is the average acceleration
of the particle over the first 20.0 seconds?
Express your answer in meters per second per second.
Hint 1. Definition and graphical interpretation of average acceleration The average acceleration of a particle that travels along a straight line in a time interval the change in velocity experienced by the particle to the time interval , or .
is the ratio of
In a v vs. t graph, then, the average acceleration equals the slope of the line connecting the two points representing the initial and final velocities.
Hint 2. Slope of a line The slope of a line from point A, of coordinates "rise" over the "run," or
, to point B, of coordinates
, is equal to the
.
ANSWER: =
0.075
Correct The average acceleration of a particle between two instants of time is the slope of the line connecting the two corresponding points in a v vs. t graph.
Part D What is the instantaneous acceleration
of the particle at
?
Hint 1. Graphical interpretation of instantaneous acceleration The acceleration of a particle at any given instant of time or at any point in its path is called the instantaneous acceleration. If the v vs. t graph of the particle's motion is known, you can directly determine the instantaneous acceleration at any point on the curve. The instantaneous acceleration at any point is equal to the slope of the line tangent to the curve at that point.
Hint 2. Slope of a line The slope of a line from point A, of coordinates "rise" over the "run," or
, to point B, of coordinates
, is equal to the
.
ANSWER: 1 0.20 =
-0.20 0.022 -0.022
Correct The instantaneous acceleration of a particle at any point on a v vs. t graph is the slope of the line tangent to the curve at that point. Since in the last 10 seconds of motion, between and , the curve is a straight line, the tangent line is the curve itself. Physically, this means that the instantaneous acceleration of the particle is constant over that time interval. This is true for any motion where velocity increases linearly with time. In the case at hand, can you think of another time interval in which the acceleration of the particle is constant?
Now that you have reviewed how to plot variables as a function of time, you can use the same technique and draw an acceleration vs. time graph, that is, the graph of (instantaneous) acceleration as a function of time. As usual in these types of graphs, time is plotted on the horizontal axis, while the vertical axis is used to indicate acceleration .
Part E Which of the graphs shown below is the correct acceleration vs. time plot for the motion described in the previous parts?
Hint 1. How to approach the problem Recall that whenever velocity increases linearly with time, acceleration is constant. In the example here, the particle's velocity increases linearly with time in the first 20.0 of motion. In the second 20.0 , the particle's velocity is const ant, and then it decreases linearly with time in the last 10 . This means that the particle's acceleration is constant over each time interval, but its value is different in each interval.
Hint 2. Find the acceleration in the first 20 What is
, the particle's acceleration in the first 20
of motion, between
and
?
Express your answer in meters per second per second.
Hint 1. Constant acceleration Since we have already determined that in the first 20 of motion the particle's acceleration is constant, its constant value will be equal to the average acceleration that you calculated in Part C.
ANSWER: =
0.075
Hint 3. Find the acceleration in the second 20
Express your answer in meters per second per second.
Hint 1. Constant velocity In the second 20 of motion, the particle's velocity remains unchanged. This means that in this time interval, the particle does not accelerate.
ANSWER: =
0
Hint 4. Find the acceleration in the last 10 What is
, the particle's acceleration in the last 10
of motion, between
and
?
Express your answer in meters per second per second.
Hint 1. Constant acceleration Since we have already determined that in the last 10 of motion the particle's acceleration is constant, its constant value will be equal to the instantaneous acceleration that you calculated in Part D.
ANSWER: =
-0.20
ANSWER: Graph A Graph B Graph C Graph D
Correct In conclusion, graphs of velocity as a function of time are a useful representation of straight-line motion. If read correctly, they can provide you with all the information you need to study the motion.
Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters.
Part A At which of the times do the two cars pass each other?
Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time.
ANSWER: A B C D E None Cannot be determined
Correct
Part B Are the two cars traveling in the same direction when they pass each other? ANSWER: yes no
Correct
Part C At which of the lettered times, if any, does car #1 momentarily stop?
Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the "rise" (change in position) over the "run" (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed.
ANSWER: A B C D E none cannot be determined
Correct
Part D At which of the lettered times, if any, does car #2 momentarily stop?
Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the "rise" (change in position) over the "run" (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed.
ANSWER: A B C D E none cannot be determined
Correct
Part E At which of the lettered times are the cars moving with nearly identical velocity ?
Hint 1. Determining Velocity from a Position versus Time Graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a
ANSWER: A B C D E None Cannot be determined
Correct
Prelecture Concept Question 2.05
Part A Suppose that you toss a rock upward so that it rises and then falls back to the earth. If the acceleration due to gravity is 9.8 m/sec 2, what is the rock’s acceleration at the instant that it reaches the top of its trajectory (where its velocity is momentarily zero)? Assume that air resistance is negligible. ANSWER: The rock has a downward acceleration of 9.8 m/s2. The rock has a downward acceleration of 19.6 m/s2. The acceleration of the rock is zero. The rock has an upward acceleration of 9.8 m/s2. The rock has an upward acceleration of 19.6 m/s2.
Correct
Problem 2.57
Part A If an object was freely falling, from what height would it need to be dropped to reach a speed of 70.0 m/s before reaching the ground? ANSWER:
322 m 250 m 712 m 500 m 189 m
Correct
Problem 2.81 A stone is thrown vertically upward with a speed of 12.0
Part A How much later does it reach the bottom of the cliff? Express your answer using three significant figures. ANSWER: = 5.32
Correct
Part B What is its speed just before hitting? Express your answer using three significant figures. ANSWER: =
40.2
Correct
from the edge of a cliff 75.0
high.
Part C What total distance did it travel? Express your answer using three significant figures. ANSWER: =
89.7
Correct
A Flea in Flight In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80 . Ignore air resistance.
Part A A flea jumps straight up to a maximum height of 0.390
. What is its initial velocity
as it leaves the ground?
Express your answer in meters per second to three significant figures.
Hint 1. Finding the knowns and unknowns Take the positive y direction to be upward, the y coordinate of the initial position of the flea to be , and denote the final height of the flea by , whose value 0.390 you know. Let be the duration of the flea's leap to its maximum height, its initial velocity, its final velocity (at maximum height), and its (constant) acceleration. Which of the following quantities is/are known? Check all that apply.
Hint 1. The number of known quantities Typically, you need to know the values of four variables in order to solve any of the kinematic equations, because they contain five variables each, with the exception of , which contains only four variables, in which case you would need to know the values of only three of these variables. Since we may place the flea at any point of the y axis to begin its jump, we have conveniently assumed that is equal to 0.
Hint 2. What is the flea's velocity at its maximum height? What is the velocity
of the flea at its maximum height of
0.390
Express your answer in meters per second to three significant figures. ANSWER: = 0
ANSWER:
?
Hint 2. Determine which kinematic equation to use Decide which kinematic equation makes the solution of this problem easiest. That is, look for an equation that contains the variable you are solving for and in which all the other variables are known. ANSWER:
Hint 3. Some algebra help You have determined that the simplest equation to use is . To solve for , you must first subtract the term from both sides of the equation, and then take the square root of both sides. Keep in mind that the acceleration is negative.
ANSWER: = 2.76
Correct
Part B How long is the flea in the air from the time it jumps to the time it hits the ground? Express your answer in seconds to three significant figures.
Hint 1. How to approach the problem One approach is to find the time it takes for the flea to go from the ground to its maximum height, and then find the time it takes for the flea to fall from its maximum height to the ground. The subsequent hints will guide you through this approach.
Hint 2. Find the time from the ground to the flea's maximum height What is the time , )?
it takes the flea to go from the ground (
Express your answer in seconds to three significant figures. ANSWER:
,
) to its maximum height (
0.390
= 0.282
Hint 3. Find the time from the flea's maximum height to the ground What is the time the ground (
that it takes for the flea to fall from its maximum height ( )?
0.390
,
) to
Express your answer in seconds to three significant figures. ANSWER: = 0.282
ANSWER: time in air =
0.564
Correct Notice that the time for the flea to rise to its maximum height is equal to the time it takes for it to fall from that height back to the ground. This is a general feature of projectile motion (any motion with ) when air resistance is neglected and the landing point is at the same height as the launch point. There is also a way to find the total time in the air in one step: just use
and realize that you are looking for the value of
for which
Score Summary: Your score on this assignment is 94.6%. You received 132.44 out of a possible total of 140 points.
.
