The effective height of the water for Smith's house will be 24.61m.
<h3>How to calculate the height?</h3>
Based on the information given, the volume of the water in sphere will be:
= 4/3πr³ = (5.80 × 10^5)/1000
= 4.18r³ = 580
r³ = 138.7
r = 5.18m
The effective height of the water will be:
= 18.0 + 2(5.18)
= 28.36
The gauge pressure at Faucet of Jones house will be:
= pgh
= 1000(9.8)(28.36)
= 277.9kPa
The effective height of the water for Smith's house will be:
= 18.0 + 2(5.18) - 3.75
= 24.61m
The gauge pressure at Faucet of Jones house will be:
= 1000 × 9.8 × 24.61
= 241.2kPa
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Answer:
0.5 , 54.5
Explanation:
for acceleration we should derivate the equation 2 times
x=3t³+t²/4
v=9t²+t/2
a=18t+1/2
a(0)=0.5
a(3)=54.5
9514 1404 393
Answer:
1.114 kg/m³
Explanation:
The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...
2910d +308 = 2910×1.22
Solving for d, we find ...
2910d = 2919(1.22) -308
d = 1.22 -308/2910
d ≈ 1.114 . . . kg/m³
The density of the hot air is about 1.114 kg/m³.
Answer:
A
Explanation:
The weight is acting downwards where as the buoyant force acting upwards (opposite) direction with equal amount of force. so the opposite forces cancel out each other (because of the force amount being equal) and no net force is acting on the object.
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