Please label answers. Thanks in advance!!! 1. Bowling balls are roughly the same
ID: 1294449 • Letter: P
Question
Please label answers. Thanks in advance!!!
1. Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.108 m, calculate the heaviest bowling ball that will float in a fluid of density 1.000 x10^3kg/m^3. Number 2. A consumer upset with the latest trend of postal rate increases has decided to try to send letters by balloon even though they may not reach their intended destination. A 72800 cm^3gas -filled balloon will provide enough lift for a 41.7 g package to be accelerated upward at a rate of 3.10 m/s^2. For these circumstances, calculate the density of the gas the consumer fills the balloon with. The acceleration due to gravity is g = 9.81 m/s^2 and the density of air is pair = 1.16 kg/m^3. Neglect the mass of the balloon material and the volume of the package. Number kg/ m^3 3' You spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.477 x10^-3m^3/s and the diameter of the nozzle you hold is 5.43 x 10^-3m. At what speed does the water exit the nozzle? Number m/sExplanation / Answer
mass of the ball = volume x density
= (4/3 pi r3) x density = (4/3 x pi x 0.1083) x 1.0 x 103
= 5.28 Kg
F = mass x acceleration
F = 5.28 kg x 9.8
F = 51.74 N
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Fb = p g V
Fb = m ay+ m g
p g V = m ay+ m g
1.16 * 9.81 * (72800 * 10-6) = m(3.10) + m(9.81)
m = 0.06417
0.06417 - 0.0417 = 0.0225 kg
density = mass / volume
0.0225 kg / (72800 * 10-6)
= 0.309 kg/m3
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speed = 0.477 x 10-3/ (pi/4 * (5.43 x 10-3)2 )
= 20.59 m/s