If a force always acts perpendicular to an object\'s direction of motion, that f

Question

If a force always acts perpendicular to an object's direction of motion, that force cannot change the object's kinetic energy. A) True B) False A small object collides with a large object and sticks. Which object experiences the larger magnitude of momentum change? A) the large object B) the small object C) Both objects experience the same magnitude of momentum change. D) cannot be determined from the information given How much work is done by 3.00 mol of ideal gas when it triples its volume at a constant temperature of 127 degree C? The ideal gas constant is R = 8.314 J/mol middot K. A) 12.7kJ B) 9.97 kJ C) 11.0 kJ D) 15.3 kJ E) 1.20 kJ An ideal gas occupies 12 liters at 293K and 1 atm (76 cm Hg). Its temperature is now raised to 373K and its pressure increased to 215 cm Hg. The new volume is: A. 0.2 liters B. 5.4 liters C. 13.6 liters D. 20.8 liters The temperature of low pressure hydrogen is reduced from 100 degree C to 20 degree C. The rms speed of its molecules decreases by approximately: A.80% B. 89% C. 46% D. 21% E. 11%

Explanation / Answer

20) True, for force in perpendicular to direction of motion it will do no work hence there is not change of kinetic energy.

Mathematically theta=90 and work done = force * distance cos theta, as cos (90)= 0 so work done is zero.

21) C, both object experience same magnitude od momentum change.

As the smaller object has smaller mass so even huge change in velocity will not change its momentum much only its mass will count, similarly the bigger object's velocity will change slight only, but since it is much more massive so small change in velocity will count more towards its momentum change.

22) C 11KJ

It is a isothermal process so

Work done = nRT*ln(Vf /Vi)

  
T = 127 C = 400K
R = 8.314 J / (mol * K)
Vf = 3*Vi

Work done= 3 * 8.314 * 400 * ln(3) = 11000 J = 11.0 kJ

23) B 5.4litre

We use Charles law
P1*V1/T1 = P2*V2/T2

now P1=1atm, T1=293K ,V1= 12litre and P2=2.82 atm, T2= 373k,V2=?

plugging values

1*12/293 = 2.82*V2/373 so V2=373*12/(293*2.82)

V2= 5.4litre

24) E 11%

Vrms = sqrt(3RT / M)
V2 / V1 =sqrt( T2 / T1)

now T2=20C=293k and T1= 373k

plugging values gives V2/V1=11%

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