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QuestionIn the triangle shown, angles ABD and DBC are 90°, AD = 15, DC = 20, and AC = 25.

What are the lengths BC and BD, respectively?

Answers12 and 16

13 and 17

16 and 12

18 and 13

The answer is (C).

SolutionFor right triangle ABD,

For right triangle DBC,

Equate the two expressions for (BD) .

Alternatively, this problem can be solved using the law of cosines.

Test BankQuestion preview

+(BD)2 (AB)2

(BD)2

= (15)2

= −(15)2 (AB)2

+(BD)2 (25 − AB)2

(BD)2

= (20)2

= −(20)2 (25 − AB)2

2

−(15)2 (AB)2

AB

BC

(BD)2

BD

= − + 50 (AB) −(20)2 (25)2 (AB)2

= = 9− +(15)2 (20)2 (25)2

50= 25 − AB = 25 − 9 = 16

= −(15)2 (9)2

= 12

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QuestionA gas mixture contains 2 moles of helium, 1 mole of neon, and equal numbers of moles of argon andkrypton. The mixture is sealed in a container with a volume of 0.03 m and is kept at three timesatmospheric pressure at a temperature of 30°C. What is most nearly the mass fraction of argon in themixture?

Answers0.08

0.16

0.19

0.39

The answer is (C).

SolutionUse the ideal gas equation to find the total number of moles in the mixture.

The number of moles of krypton is equal to the number of moles of argon.

Find the mole fraction of each of the gases in the mixture.

Test BankQuestion preview

3

Nmixture =pV

TR̄̄̄

=

(3) (101.3 kPa) (0.03  )(1000  )m3 mol

kmol

(8.314  ) (30°C + 273°)kJ

kmol⋅K

= 3.6 mol

NKr = =NAr− +Nmixture NHe NNe

2

=3.6 mol − 2 mol − 1 mol

2= 0.3 mol

QUESTION DATAVendor0000087622Solving Time Difficultyeasy Quantitative?No StatusActiveCreated On07/11/2018 09:28:15PMPublished On07/11/2018 09:28:15PMModified On01/03/2020 08:32:53PMOTHER VERSIONS

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Solve for the mass fraction of argon in the mixture.

xHe

xNe

xKr

=NHe

Nmixture

=2 mol

3.6 mol= 0.553

=NNe

Nmixture

=1 mol

3.6 mol= 0.276

= =xArNAr

Nmixture

=0.3 mol

3.6 mol= 0.086

yAr =xArMAr

+ + +xHeMHe xNeMNe xArMAr xKrMKr

=(0.086)(40  )

g

mol

(0.553)(4  ) + (0.276)(20  )g

mol

g

mol

+ (0.086)(40  ) + (0.086)(84  )g

mol

g

mol= 0.187  (0.19)

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QuestionA bare, horizontal conductor with a circular cross section and an outside diameter of 1.5 cm dissipates25 W per meter of wire length. The conductor is cooled by free convection, and the surrounding airtemperature is 15°C. The film temperature is 38°C. The film coefficient is 10.27 W/m ·K. Most nearly,what is the conductor’s surface temperature?

Answers29°C

43°C

54°C

67°C

The answer is (D).

SolutionThe wire diameter is

The heat transfer from the wire is

The surface temperature is

Test BankQuestion preview

2

d = = 0.015 m1.5 cm

100 cm

m

Q⋅

= hA( − ) = hπdL ( − )Ts T∞ Ts T∞

Ts = +

Q⋅

L

hπdT∞

= + 15°C

25 W

1 m

(10.27  )π (0.015 m)W

⋅Km2

= 66.7°C (67°C)

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Question

A 2 kg block rests on a 34° incline.

The coefficient of static friction is 0.2. Approximately how much additional force, F, must be applied to keep the block from sliding down the incline?

Answers

(A) 7.7 N

(B) 8.8 N

(C) 9.1 N

(D) 14 N

The answer is (A).

Solution

Choose coordinate axes parallel and perpendicular to the incline.

The sum of the forces is

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~ Fx = 0 = F + Ft - W sin¢,

F = Wsin¢,- Ft

= mgsin¢,- µ 8 N

= mgsin¢,- µ 8 mgcos¢,

= mg (sin¢, - µ cos ¢,)

= (2 kg) ( 9.81 : ) (sin34 ° - 0.2cos34 °)

= 7.7 N (7.7 N)

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QuestionA square column with a solid cross section is placed in a building to support a load of 5 MN. Themaximum allowable stress in the column is 350 MPa. The column reacts linearly to all loads. If thecontractor is permitted to load the column anywhere in the central one-fifth of the column’s cross section,what are most nearly the smallest possible dimensions of the column?

Answers12 cm × 12 cm

14 cm × 14 cm

16 cm × 16 cm

18 cm × 18 cm

The answer is (D).

SolutionThe middle one-fifth of the column is a square with dimensions of b/5 × b/5 (0.2b × 0.2b).

The maximum stress will be induced when the middle one-fifth square is loaded at one of its corners.

The cross-sectional area is

The moment of inertia of the square cross section is

The distance from the neutral axis to the extreme fibers is

Test Bank Question preview

A = b2

I =b4

12

c =b

2

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The maximum eccentricity is

The stress at the extreme corner is

e = 0.1b

σ

 

b

= ± + = F ±F

A

Fexcx

Ix

Feycy

Iy

⎛

⎝

⎜⎜⎜1

b2

(2) (0.1b)( )b

2

b4

12

⎞

⎠

⎟⎟⎟

= F ( ± )1

b2

1.2

b2

=2.2F

b2

= =2.2F

σ

− −−−−√

(2.2) (5 MN)(   )106 N

MN

(350 MPa)(   )106 Pa

MPa

− −−−−−−−−−−−−−−−−−−−

⎷

= 0.177 m (18 cm)

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QuestionA pump with a mass of 45 kg is supported by four springs, each having a spring constant of 1750 N/m.The motor is constrained to allow only vertical movement. The natural frequency of the pump is mostnearly

Answers6.0 rad/s

9.0 rad/s

12 rad/s

15 rad/s

The answer is (C).

SolutionCalculate the total spring constant.

Calculate the static deflection.

Calculate the natural frequency.

Test BankQuestion preview

k = (4)(1750  ) = 7000 N/mN

m

mg

δst

= kδst

= =mg

k

(45 kg)(9.81  )m

s2

7000 N

m= 0.063 m

ω = =g

δst

−−−√

9.81 m

s2

0.063 m

− −−−−−−

⎷

= 12.48 rad/s (12 rad/s)

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QuestionA block diagram is shown.

G = –5 dB, G = 2 dB, G = 4 dB, and G = 3 dB. What is the system sensitivity?

Answers0.15

0.37

1.72

2.25

The answer is (C).

SolutionSimplify the block diagram.

Test BankQuestion preview

1 2 3 4

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Convert the block gains from dB to ratios.

From step 2 of the simplified block diagram,

The system has positive feedback. The sensitivity of the system is

= = 0.316G1 10− 510/

= = 1.585G2 102

10/

= = 2.512G3 104

10/

= = 1.995G4 103

10/

G (s)

G (s)H(s)

= +G1

1 + G1G2

G3

= 0.316

1+(0.316)(1.585)

= 0.2105= 1.995

= (0.2105)(1.995)

= 0.4200

S =1

1 − G(s)H(s)

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=1

1 − 0.4200

= 1.724 (1.72)