Q. 9 : Amount of energy required to blow a bubble of radius 5 cm, (surface tension of soap is 30 x 10-2 N/m) (a) 1.88 J (b) 1.88 х 10-1] (c) 1.88 × 10-² J (d) 1.88 x 10 J
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A: Mass of plant (m) = 0.5 kg Original gravitational potential energy (E) = 70 Joules
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A: length (L) = 1.07 m mass (m) = 4.92 kg initial speed = 2.88 msenergy lost = 1 J
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A: Power, P=200 WTime , t=1 hr=3600 sLv=2.26×106 J/kgenergy used. e=15%
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A: radius = 2.1 km h = 2.7 cm density of water = 1000 kg/m3
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A: Given: Unit conversions are as follows: 1kg=1000g 1m=100cm
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Q: Woukdone in blocuing a soap bubble of diamelere 2 cm is (S.T. = = 3 × 10 ² N/mm). (a) 7-54 x 10-5 J…
A: We have to compute-Work done in blowing a soap bubble (W)=?Given that-Surface tension (T)=3×10-2…
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A: To find-Energy required (E)=?Given-S.T. (T)=0.035 N/mr1=4 cm=4×10-2 mr2=6 cm=6×10-2 m
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Q: This time, the pendulum is 1.89 m long and has a mass of 3.24 kg. You give it a push away from…
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A: From ideal gas law,Thus, we can find the internal energy to be,
Q: Wouk done in blocuing a soap bubble of diameter 2 cm is (S.T. =3×102 N/mm) (a) 7-54 x 10-5 J (b)…
A: We have to compute-Work done in blowing a soap bubble (W)=?Given that-Surface tension (T)=3×10-2…
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Q: 5. This time, the pendulum is 1.55 m long and has a mass of 3.07 kg. You give it a push away from…
A: mass of the pendulum, m=3.07 kg length of the pendulum, L=1.55 m initial velocity applied, v=2.65…
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A: Mass, m = 3.2 g Radius, r = h = 19 cm = 0.19 m Initial speed, vi = 0.51 m/s
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Q: Ex. 2: There is soap film on a rectangular frame the process ? (Surface tension of soap film of wire…
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