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:
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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Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
