Answer:
The minimum wall thickness required for the spherical tank is 0.0189 m
Explanation:
Given data:
d = inside diameter = 8.1 m
P = internal pressure = 1.26 MPa
σ = 270 MPa
factor of safety = 2
Question: Determine the minimum wall thickness required for the spherical tank, tmin = ?
The allow factor of safety:
The minimun wall thickness:
Answer:
0.31 m
Explanation:
m = mass of the block = 1.5 kg
H = height from which the block is released on ramp = 0.81 m
k = spring constant of the spring = 250 N/m
x = maximum compression of the spring
using conservation of energy
Spring potential energy gained by spring = Potential energy lost by block
(0.5) k x² = mgH
(0.5) (250) x² = (1.5) (9.8) (0.81)
x = 0.31 m
The change in potential energy when the block falls to ground is -480J.
The maximum change in kinetic energy of the ball is 480 J.
The initial kinetic energy of the ball is 0 J.
The final kinetic energy of the ball is 0.148J.
The initial potential energy of the ball is 0.187 J.
The final potential energy of the ball is 0 J.
The work done by the air resistance is 0.039 J.
<h3>Change in potential energy when the block falls to ground</h3>
ΔP.E = -mgh
ΔP.E = -Wh
ΔP.E = - 40 x 12
ΔP.E = -480 J
<h3>Maximum change in kinetic energy of the ball</h3>
ΔK.E = - ΔP.E
ΔK.E = - (-480 J)
ΔK.E = 480 J
<h3>Initial kinetic energy of the ball</h3>
K.Ei = 0.5mv²
where;
- v is zero since it is initially at rest
K.Ei = 0.5m(0) = 0
<h3>Final kinetic energy</h3>
K.Ef = 0.5mv²
K.Ef = 0.5(0.0091)(5.7)²
K.Ef = 0.148 J
<h3>Initial potential energy of the ball</h3>
P.Ei = mghi
P.Ei = 0.0091 x 9.8 x 2.1
P.Ei = 0.187 J
<h3>Final potential energy</h3>
P.Ef = mghf
P.Ef = 0.0091 x 9.8 x 0
P.Ef = 0
<h3>Work done by the air resistance</h3>
W = ΔE
W = P.E - K.E
W = 0.187 J - 0.148 J
W = 0.039 J
Learn more about potential energy here: brainly.com/question/1242059
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Answer:
The answer to your question is 1800 Pa
Explanation:
Data
Weight = 2700 N
Area = 1.5 m²
Pressure = ?
Formula
Pressure = Force / Area
The Pressure is defined as the force exerted per unit area.
-Substitution
Pressure = 2700 / 1.5
-Result
Pressure = 1800 Pa
The units of pressure are Pascals (Pa)
