An 1800-kg elephant is placed on top of an aluminum cylinder, 4.3 meters high. The area of the top of the cylinder is 0.080 m². By how much does the weight of the elephant compress the cylinder? Young's modulus for aluminum is Y = 7.0 X 1010 N/m?. Me = 1800 kg A = 0.080 m2 4.3 m

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Chapter1: Units, Trigonometry. And Vectors
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Young's modulus
F
AL
=Y
A
L.
Density:
M
(kg /m³)
p=.
V
F
P =
A
P = P, + pgh
Pressure:
(Ра)
Buoyancy Force, B = weight in air – weight in fluid
Area of a circle = A = Tr²
weight = mg,
where
g
= 9.80 m/s?
Pressure:
F
P=
(Ра)
A
In liquid: P= P,+pgh
Buoyancy Force, B = weight in air – weight in fluid
Area of a circle = A = Tr?
Buoyancy force on a submerged object: B
Pfluid Vokiect g
Fluid flow:
Flow rate Q(m³/s)
VA
VIA1 = V2A2
Bernoulli's Principle: P1 + pghi + ½ pvi? = P2+ pgh2 + ½ p vz?
Transcribed Image Text:Young's modulus F AL =Y A L. Density: M (kg /m³) p=. V F P = A P = P, + pgh Pressure: (Ра) Buoyancy Force, B = weight in air – weight in fluid Area of a circle = A = Tr² weight = mg, where g = 9.80 m/s? Pressure: F P= (Ра) A In liquid: P= P,+pgh Buoyancy Force, B = weight in air – weight in fluid Area of a circle = A = Tr? Buoyancy force on a submerged object: B Pfluid Vokiect g Fluid flow: Flow rate Q(m³/s) VA VIA1 = V2A2 Bernoulli's Principle: P1 + pghi + ½ pvi? = P2+ pgh2 + ½ p vz?
An 1800-kg elephant is placed on top of an aluminum cylinder, 4.3 meters high. The
area of the top of the cylinder is 0.080 m². By how much does the weight of the
elephant compress the cylinder? Young's modulus for aluminum is Y = 7.0 X 1010
N/m?.
Me= 1800 kg
A = 0.080 m2
4.3 m
Transcribed Image Text:An 1800-kg elephant is placed on top of an aluminum cylinder, 4.3 meters high. The area of the top of the cylinder is 0.080 m². By how much does the weight of the elephant compress the cylinder? Young's modulus for aluminum is Y = 7.0 X 1010 N/m?. Me= 1800 kg A = 0.080 m2 4.3 m
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