Suppose the ends of a 14 m long steel beam are rigidly clamped at 0 degree C to

Question

Suppose the ends of a 14 m long steel beam are rigidly clamped at 0 degree C to prevent expansion. The cm^2. What force does the beam exert when it is heated to 45 degree C? (alpha-alpha_ = 1.1 times 10^C degree, Y_ = 2 times 10^N/m^2) A) 3740 N B) 630 N C) 5208 N D) 11880 N E) 17640 N A vertical spring with a spring constant 46 N/m is hanging from a ceiling. A small object with a 1.7 kg/s then attached. The object is then pulled down and released. The speed of the object when it is 10 cm from the equilibrium position is 1.9 m/s. How far down was the object pulled? A) 41.2 cm B) 87.4 cm C) 21.896 cm D) 37.87 cm E) 27.968 cm The density of a certain metal solid is 6.8 times 10^3 kg/m^3 and its Young's modulus is 0.9 times 10^7. What is the velocity of sized is this metal? A) 1958.4 m/s B) 3060 m/s C) 3638 m/s D) 2312 m/s E) 1428 m/s A gasoline engine with an efficiency of 22% operates between a high temperature T_1 and a low temperature T_1 = 20 degree C. If this engine operates with Carnot efficiency, what is the high-side temperature T_1? A) 194.92 degree C B) 59.4 degree C C) 489 degree C D) 102.6 degree C E) 19.8 degree C Diatomic hydrogen explodes adiabatically to 36 times its original volume before its pressure reduces back to atmospheric pressure. When ignited, what was its initial pressure? A) 150.9 Atm B) 216 Atm C) 77.4 Atm D) 90 Atm E) 7.2 Atm What is the first overtone frequency for a 2 m length organ pipe, closed at one end? The speed of sound is air is 333 m/s? A) 124.9 Hz B) 52.2 Hz C) 76.6 Hz D) 146.6 Hz E) 19.98 Hz

Explanation / Answer

5. length should be as temperature increases.

L = L0 (1 + alpha deltaT) = 14 (1 + (1.1 x 10^-6)(45)) = 14.000693 m

F / A = Y (deltaL/ L)

F / (12 x 10^-4) = (2 x 10^11) (693 x 10^-6) / 14

F = 11880 N

Ans(D)

6. total energy = k x^2 /2 + m v^2 /2 = k A^2 /2

46(0.10^2) + (1.7)(1.9^2) = 46 A^2

A = 0.3787 m Or 37.87 cm

Ans(D)

7. v = sqtr[ Y / rho]

= sqrt[ (0.9 x 10^11) / 6.8 x 10^3] = 3638 m/s

Ans(C)

8. n = 0.22 = 1 - (T_cold / T_hot)

(273 + 20)/(273 + T) = 1- 0.22 = 0.78

T = 102.6 deg C

Ans(D)

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