Phy 121 Ch 13-15 Exam Name



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Alternate Exam



The motion of the piston of an automobile engine is approximately simple harmonic.

1. If the stroke (twice the amplitude) of an engine is 0.100m and the engine runs at 4000rpm, compute the acceleration of the piston at the endpoint of its stroke.

2. If the piston has a mass of 500 grams, what net force must be exerted on it at this point?



3. What is the kinetic energy of the piston at the midpoint of its stroke? (max values of sin or cos are 1)




6/6 6/6

a = A ω2 cos(ω t)

a = 0.05(2π4000/60)2

a = 8760 m/s2



6/6

F = m a


F = .5 (8760)

F = 4380 N



6/6

v = A ω sin(ω t)

v = 0.05(2π4000/60)

v = 20.9 m/s



8/8

K = ½ m v2

K = ½ (.5Kg) 20.92

K = 110 Joules









Or Kmax = ½ m vmax2

Kmax = ½ m ω2 A2

Kmax = ½ k A2



An organ pipe 5.0m long is open at one end, similar to your lab. What is the linear distance between a node and the adjacent antinode for the 5th harmonic in this organ pipe?

(Hint, this is easier if you draw in the nodes/antinodes)

5λ/4 = 5 m 10/10



λ = 4 meters

N to A is λ / 4 10/10



N to A = 4 / 4 = 1 meter

12/12





A 200 gram ball with a volume of 100 ml is lowered on a string into a container with two different fluids. If the tension in the string is ½ Newton when the ball is half submerged in the more dense fluid (1.8 g/cc) what is the density of the less dense fluid?

Fup = Fdown since FNet = mt a  a = zero

FT + FB-Bottom half + FB-Top half = FW

3/3 10/10 10/10 8/8 3/3

0.5 N + 1800(50x10-6)g + ρ(50x10-6)g = 0.2 kg “g”

0.5 N + 0.9 + ρ (0.0005) = 2 N

ρ = 1200 kg/m3





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