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:
1 m/s² down
Explanation:
"Calculate the acceleration of the elevator, and find the direction of acceleration."
Sum of forces in the +y direction:
∑F = ma
N − mg = ma
0.9 mg − mg = ma
-0.1 mg = ma
a = -0.1 g
a = -1 m/s²
Answer:
i think it might be c but i know its not A or B and im
on the fence for D so the best answer would be C
Explanation:
Answer:
3000N
Explanation:
divided to get answer
the force needed to accelerate the 1000kg car by 3m/s2 is 3000N
The car will take 300 m before it stops due to applying break.
<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
- As per Newton's equation of motion, V² - U² = 2aS
- V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
- Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
- So, 0² - 60² = 2×6× S
=> -3600 = -12S
=> S = 3600/12 = 300 m
Thus, we can conclude that the distance covered by the car is 300 m before it stopped.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?
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