Consider the double-fluid manometer attached to an air pipe shown. If the specific gravity of one fluid is 14.514.5, determine t
he specific gravity of the other fluid if pressure of air was 81kpa81kpa. Take the atmospheric pressure to be 100 kpa.
1 answer:
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is
Explanation:
From the question we are told that
The specific gravity of the first fluid is
The pressure of air is
The atmospheric pressure is
Generally the specific gravity of the second fluid is mathematically represented as
Here is the density of water with value
=>
=>
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Answer:
Explanation:
a ) Time period T = 2 s
Angular velocity ω = 2π / T
= 2π / 2 = 3.14 rad /s
Initial moment of inertia I₁ = 200 + mr²
= 200 + 25 x 2.5²
=356.25
Final moment of inertia
I₂ = 200 + 25 X 1.5 X 1.5
= 256.25
b ) We apply law of conservation of momentum
I₁ X ω₁ = I₂ X ω₂
ω₂ = I₁ X ω₁ / I₂
Putting the values
ω₂ = 4.365 rad s⁻¹
c ) Increase in rotational kinetic energy
=1/2 I₂ X ω₂² - 1/2 I₁ X ω₁²
.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²
= 684.95 J
This energy comes from work done against the centripetal pseudo -force.
Answer:
a)1.37 s
b)∞ ( Infinite)
Explanation:
Given that
L= 47 cm ( 1 m =100 cm)
L= 0.47 m
a)
On the earth :
Acceleration due to gravity = g
We know that time period of the simple pendulum given as
Here
Now by putting the values
T=1.37 s
b)
Free falling elevator :
When elevator is falling freely then
( This is case of weightless motion)
Therefore
T=∞ (Infinite)