A heat engine absorbs 150 J of heat from a hot reservoir and rejects 90 J to a c

Question

A heat engine absorbs 150 J of heat from a hot reservoir and rejects 90 J to a cold reservoir. What is the efficiency of this engine? A) 20% B) 40% C) 60% D) 67% E) 90% The maximum theoretical thermal efficiency of a steam engine that is supplied steam at a temperature of 600 degree C and exhausts it at a temperature of 200 degree C is A) 33.3% B) 45.8% C) 66.7% D) 77.1% E) 84.6% The coefficient of linear expansion of the steel in a certain bridge is 1.10 times 10^-5/C degree. If at a minimum winter temperature of -30 degree C the bridge is 30.5 m long, at the summer maximum of 40 degree C its length is increased about A) 2.54 cm B) 2.54 mm C) 0.254 mm D) 25.4 mum E) 2.54 mum If the thickness of a uniform wall is doubled, the rate at which heat is conducted through the wall is A) doubled. B) increased by a factor of four. C) decreased by a factor of four. D) cut in half. E) unchanged. Two small spheres, each with mass m = 5.0 g and charge q, are suspended from a point by threads of length L = 0.30 m. What is the charge on each sphere if the threads make an angle theta = 20 degree with the vertical? A) 7.9 times 10^-7 C D) 6.3 times 10^13 C B) 2.9 times 10^-7 C E) 1.8 times 10^-7C C) 7.5 times 10^-2 C Point charges of 4.0 times 10^-8 C and -2.0 times 10^-8 C are placed 12 cm apart. A third point charge of 3.0 times 10^-8 C halfway between the first two point charges experiences a force of magnitude A) 4.5 times 10^-3 N B) 2.0 times 10^-3 N C) 1.5 times 10^-3 N D) zero E) 5.0 times 10^-3 N

Explanation / Answer

14) n = (Qh - Qc)/Qh

n = (150-50)/150 = 0.67

n = 67%

correct option is (D)

15) n = (Th-Tc)/Th

n = (600-200)/(600+273) = 0.458

n = 45.8%

correct option is (B)

16) alpha = 1.1*10^-6 /Co

T1 = -30 oC, T2 = 40 oC , L = 30.5 m

alpha = DL/(T2 -T1)

1.1*10^-6 = DL/(30.5*(40+30))

DL = 0.00235 m

DL = 2.35 mm

correct piton is (B)

17) H1= Q/t = KA*(T1-T2)/x

X2 = 2x1

H2 = H1/2

correct otpion is(D)

18) Tcos(20) = mg

Tcos(20) = 0.005*9.8

T = 0.052 N

cos(20) = x/L

cos(20) = x/0.3

x = 0.282

r = 0.282*2 = 0.564 m

Tsin(20) = kq1q2/r^2

0.052*sin(20) = 9*10^9*q^2/0.564^2

q = 7.9*10^-7 C

correct option is (A)

19) F = F1+F2

F = (9*10^9*4*10^-8*3*10^-8/0.06^2) + (9*10^9*2*10^-8*3*10^-8/0.06^2)

F= 4.5*10^-3 N

correct option is (A)

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