Calculus: Early Transcendentals
4th Edition
ISBN: 9781319050740
Author: Jon Rogawski, Colin Adams, Robert Franzosa
Publisher: W. H. Freeman
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Chapter 7.6, Problem 34E
To determine
Evaluate the
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Question 1:
If a barometer were built using oil (p = 0.92 g/cm³) instead of mercury (p =
13.6 g/cm³), would the column of oil be higher than, lower than, or the same as the
column of mercury at 1.00 atm? If the level is different, by what factor? Explain. (5 pts)
Solution:
A barometer works based on the principle that the pressure exerted by the liquid column
balances atmospheric pressure. The pressure is given by:
P = pgh
Since the atmospheric pressure remains constant (P = 1.00 atm), the height of the
liquid column is inversely proportional to its density:
Step 1: Given Data
PHg
hol=hgx
Poil
• Density of mercury: PHg = 13.6 g/cm³
Density of oil: Poil = 0.92 g/cm³
• Standard height of mercury at 1.00 atm: hμg
Step 2: Compute Height of Oil
= 760 mm = 0.760 m
13.6
hoil
= 0.760 x
0.92
hoil
= 0.760 × 14.78
hoil
= 11.23 m
Step 3: Compare Heights
Since oil is less dense than mercury, the column of oil must be much taller than that of
mercury. The factor by which it is taller is:
Final…
Question 3:
A sealed flask at room temperature contains a mixture of neon (Ne) and nitrogen (N2)
gases.
Ne has a mass of 3.25 g and exerts a pressure of 48.2 torr.
.
N2 contributes a pressure of 142 torr.
• What is the mass of the N2 in the flask?
• Atomic mass of Ne = 20.1797 g/mol
• Atomic mass of N = 14.0067 g/mol
Solution:
We will use the Ideal Gas Law to determine the number of moles of each gas and
calculate the mass of N2.
PV = nRT
where:
• P = total pressure
•
V volume of the flask (same for both gases)
n = number of moles of gas
•
R 0.0821 L atm/mol K
• T = Room temperature (assume 298 K)
Since both gases are in the same flask, their partial pressures correspond to their mole
fractions.
Step 1: Convert Pressures to Atmospheres
48.2
PNe
= 0.0634 atm
760
142
PN2
=
= 0.1868 atm
760
Step 2: Determine Moles of Ne
nNe =
mass
molar mass
3.25
nNe
20.1797
nne 0.1611 mol
Step 3: Use Partial Pressure Ratio to Find n
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Chapter 7 Solutions
Calculus: Early Transcendentals
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- 3.12 (B). A horizontal beam AB is 4 m long and of constant flexural rigidity. It is rigidly built-in at the left-hand end A and simply supported on a non-yielding support at the right-hand end B. The beam carries Uniformly distributed vertical loading of 18 kN/m over its whole length, together with a vertical downward load of 10KN at 2.5 m from the end A. Sketch the S.F. and B.M. diagrams for the beam, indicating all main values. Cl. Struct. E.] CS.F. 45,10,376 KN, B.M. 186, +36.15 kNm.7arrow_forwardQize f(x) = x + 2x2 - 2 x² + 4x²² - Solve the equation using Newton Raphsonarrow_forward-b±√√b2-4ac 2a @4x²-12x+9=0 27 de febrero de 2025 -b±√√b2-4ac 2a ⑥2x²-4x-1=0 a = 4 b=-12 c=9 a = 2 b = 9 c = \ x=-42±√(2-4 (4) (9) 2(4)) X = (12) ±√44)-(360) 2(108) x = ±√ X = =±√√²-4(2) (1) 2() X = ±√ + X = X = + X₁ = = X₁ = X₁ = + X₁ = = =arrow_forward
- 3.9 (A/B). A beam ABCDE, with A on the left, is 7 m long and is simply supported at Band E. The lengths of the various portions are AB 1-5m, BC = 1-5m, CD = 1 m and DE : 3 m. There is a uniformly distributed load of 15kN/m between B and a point 2m to the right of B and concentrated loads of 20 KN act at 4 and 0 with one of 50 KN at C. (a) Draw the S.F. diagrams and hence determine the position from A at which the S.F. is zero. (b) Determine the value of the B.M. at this point. (c) Sketch the B.M. diagram approximately to scale, quoting the principal values. [3.32 m, 69.8 KNm, 0, 30, 69.1, 68.1, 0 kNm.]arrow_forward4. Verify that V X (aẢ) = (Va) XẢ + aV X Ả where Ả = xyz(x + y + 2) A and a = 3xy + 4zx by carrying out the detailed differentiations.arrow_forward3. For each of the arrow or quiver graphs shown below, determine analytically V°C and V X Č. From these analytical solutions, identify the extrema (+/-) and plot these points on the arrow graph. (a) C = −✰CosxSiny + ŷSinxCosy -π<ׂу<π Ty (b) C = −xSin2y + ŷCos2y x, y<π -π< (c) C = −xCosx + ŷSiny -π< x, y < πarrow_forward
- 7.10 (B/C). A circular flat plate of diameter 305 mm and thickness 6.35 mm is clamped at the edges and subjected to a Uniform lateral pressure of 345 kN/m². Evaluate: (a) the central deflection, (b) the position and magnitude of the maximum radial stress. C6.1 x 10 m; 149.2 MN/m².] 100 200arrow_forward3.15 (B). A beam ABCD is simply supported at B and C with ABCD=2m; BC 4 m. It carries a point load of 60 KN at the free end A, a Uniformly distributed load of 60 KN/m between B and C and an anticlockwise moment of 80 KN m in the plane of the beam applied at the free end D. Sketch and dimension the S.F. and B.M. diagrams, and determine the position and magnitude of the maximum bending moment. CEL.E.] CS.F. 60, 170, 70KN, B.M. 120, +120.1, +80 kNm, 120.1 kNm at 2.83 m to right of 8.7arrow_forward7.1 (A/B). A Uniform I-section beam has flanges 150 mm wide by 8 mm thick and a web 180 mm wide and 8 mm thick. At a certain section there is a shearing force of 120 KN. Draw a diagram to illustrate the distribution of shear stress across the section as a result of bending. What is the maximum shear stress? [86.7 MN/m².arrow_forward
- 1. Let Ả = −2x + 3y+42, B = - - 7x +lý +22, and C = −1x + 2y + 42. Find (a) Ả X B (b) ẢX B°C c) →→ Ả B X C d) ẢB°C e) ẢX B XC.arrow_forward3.13 (B). A beam ABC, 6 m long, is simply-supported at the left-hand end A and at B I'm from the right-hand end C. The beam is of weight 100 N/metre run. (a) Determine the reactions at A and B. (b) Construct to scales of 20 mm = 1 m and 20 mm = 100 N, the shearing-force diagram for the beam, indicating thereon the principal values. (c) Determine the magnitude and position of the maximum bending moment. (You may, if you so wish, deduce the answers from the shearing force diagram without constructing a full or partial bending-moment diagram.) [C.G.] C240 N, 360 N, 288 Nm, 2.4 m from A.]arrow_forward5. Using parentheses make sense of the expression V · VXVV · Å where Ả = Ã(x, y, z). Is the result a vector or a scaler?arrow_forward
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