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14 Nov 2019

What's your normal body temperature? It may not be 98.6°F,theoft-quoted average that was determined in the nineteenthcentury. Amore recent study has reported an average temperature of98.2°F.What is the difference between these averages,expressed inCelsius degrees?

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Tod Thiel
Tod ThielLv2
23 Oct 2019

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

The kinetic theory of gases states that the kinetic energy of a gas is directly proportional to the temperature of the gas. A relationship between the microscopic properties of the gas molecules and the macroscopic properties of the gas can be derived using the following assumptions: The gas is composed of pointlike particles separated by comparatively large distances. The gas molecules are in continual random motion with collisions being perfectly elastic. The gas molecules exert no long-range forces on each other. One of the most important microscopic properties of gas molecules is velocity. There are several different ways to describe statistically the average velocity of a molecule in a gas. The most obvious measure is the average velocity Vavg. However, since the molecules in a gas are moving in random directions, the average velocity is approximately zero. Another measure of velocity is (V^2)avg, the average squared velocity. Since the square of velocity is always positive, this measure does not average to zero over the entire gas. A third measure is the root-mean-square (rms) speed,Vrms , equal to the square root of (V^2)avg. The rms speed is a good approximation of the the typical speed of the molecules in a gas. This histogram (Figure 1) shows a theoretical distribution of speeds of molecules in a sample of nitrogen () gas. In this problem, you'll use the histogram to compute properties of the gas.

 

Part A
What is the average speedv_avgof the molecules in the gas?
Part B

Because the kinetic energy of a single molecule is related to its velocity squared, the best measure of the kinetic energy of the entire gas is obtained by computing the mean squared velocity,(v^2)_{\rm avg}, or its square rootv_rms. The quantityv_rmsis more common than(v^2)_{\rm avg}because it has the dimensions of velocity instead of the less-familiar velocity-squared.

What is the rms speedv_rmsof the molecules in the nitrogen gas?

Express your answer numerically to three significant digits.
Part C

What is the temperatureTof the sample of\rm N_2gas described in the histogram?

Express your answer in degrees Celsius to three significant figues.

A magnetic field of 0.1T forces a proton beam of 1.5 mA to move ina circle of radius 0.1 m. The plane of the circle is perpendicularto the magnetic field.

47. Of the following, which is the best estimate of the speed of aproton in the beam as it moves in the circle?
(A) 10-2 m/s (B) 103 m/s (C) 106 m/s (D) 108 m/s (E) 1015 m/s

52. An ideal gas is initially in a state that corresponds to point1 on the graph above, where it has pres¬sure p1, volume V1, andtemperature T1. The gas undergoes an isothermal process representedby the curve shown, which takes it to a final state 3 attemperature T3. If T2 and T4 are the tempera¬tures the gas wouldhave at points 2 and 4, respec¬tively, which of the followingrelationships is true?
(A) T1 < T3 (B) T1 < T2 (C) T1 < T4 (D) T1 = T2 (E) T1 =T4

53. The absolute temperature of a sample of monatomic ideal gas isdoubled at constant volume. What effect, if any, does this have onthe pressure and density of the sample of gas?
Pressure Density
(A) Remains the same Remains the same
(B) Remains the same Doubles
(C) Doubles Remains the same
(D) Doubles Is multiplied by a factor of 4
(E) Is multiplied by a factor of 4 Doubles



62. A sample of an ideal gas is in a tank of constant volume. Thesample absorbs heat energy so that its temperature changes from 300K to 600 K. If v1 is the average speed of the gas molecules beforethe absorption of heat and v2 is their average speed after theabsorption of heat, what is the ratio v2/ v1 ?
(A) 1/2 (B) 1 (C) (D) 2 (E) 4

63. Two people of unequal mass are initially standing still on icewith negligible friction. They then simul¬taneously push each otherhorizontally. Afterward, which of the following is true?
(A) The kinetic energies of the two people are equal.
(B) The speeds of the two people are equal.
(C) The momenta of the two people are of equal magnitude.
(D) The center of mass of the two person system moves in thedirection of the less massive person.
(E) The less massive person has a smaller initial acceleration thanthe more massive person.

64. Two parallel conducting plates, separated by a distance d, areconnected to a battery of emf e. Which of the following is correctif the plate separa¬tion is doubled while the battery remainsconnected?
(A) The electric charge on the plates is doubled.
(B) The electric charge on the plates is halved.
(C) The potential difference between the plates is doubled.
(D) The potential difference between the plates is halved
(E) The capacitance is unchanged.

68. A rod on a horizontal tabletop is pivoted at one end and isfree to rotate without friction about a vertical axis, as shownabove. A force F is applied at the other end, at an angle ? to therod. If F were to be applied perpendicular to the rod, at whatdistance from the axis should it be applied in order to pro¬ducethe same torque?
(A) L sin ? (B) L cos ? (C) L (D) L tan ? (E) L

70. A 4 µF capacitor is charged to a potential differ¬ence of 100V. The electrical energy stored in the capacitor is
(A) 2 x 10-10 J (B) 2 x 10 8 J (C) 2 x 10 6 J (D) 2x 10-4 J (E) 2 x10-2 J



Could you please proofread and correct scientifically andgrammatically my work, thanks ^_^

Here itis:
The percentage error was found to be 0.186% and 6.66%respectively.

This implies that the predicted value of the freefall accelerationis fairly accurate while the predicated value of deceleration isnot as much. The encountered errors were due to severalexperimental and human limitation was that the experiment involved.The experimental limitation was due to the uncertainty of theclinometer which was ± 0.5o.

I.e. In our calculation, a difference of 0.5o can lead to adifference of 40 - 80 cm in height as shown below:

The human limitations was due to the human eye being unable todetect the exact point on the track at which the gondoladecelerates. This is because the braking happened very quickly.This adds an uncertainty to the location of deceleration when theangels are being measured.

In addition, the human ability cannot decide on the exact period oftime at which the gondola freefalls or decelerates. This is due tothe human reaction speed being unable to exactly determine thestart time of the momentary freefall and deceleration. The timetaken for humans to react cannot be measured thus it cannot besubtracted from the findings to get an accurate timing.

All of these limitations that have affected the accuracy of theresults could be avoided to some extent by marking the exact zonesrequired on the track of the giant drop as well as using anelectronic angle measurer to measure the angles . Additionally, theusage of an electronic timer to measure periods of the freefall anddeceleration is also suggested.

During the analysis, finding an average to the deceleration in theaccelerometer reading (using Logger Pro) was only estimated usingthe human naked eye. This is because an average could not be takenelectronically. Thus, there is also some uncertainty involved inthe theoretical value.

NOTE:

The value for freefall acceleration obtained from thevideo-analysis technique was 9.754 ± 0.3237 m/s2. 9.754 m/s2 has apercentage error of :
0.469%

This value indicates it is fairly accurate and can be used as aconforming result to the theoretical value. The error involved isdue to the fact that the video was shaking, thus plotting thevelocity/ time graph of the gondola at different times wasnegatively affected; although the axes was roughly re-set at everyframe.

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