If the light strikes the plastic (from the water) at an angle θw, at what angle θa does it emerge from the plastic (into the air
2 answers:
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Answer:
2.06 m³/s
Explanation:
diameter of pipe, d = 0.81 m
diameter of constriction, d' = 0.486 m
radius, r = 0.405 m
r' = 0.243 m
density of oil, ρ = 821 kg/m³
Pressure in the pipe, P = 7970 N/m²
Pressure at the constriction, P' = 5977.5 N/m²
Let v and v' is the velocity of fluid in the pipe and at the constriction.
By use of the equation of continuity
A x v = A' x v'
r² x v = r'² x v'
0.405 x 0.405 x v = 0.243 x 0.243 x v'
v = 0.36 v' .... (1)
Use of Bernoulli's theorem
7970 + 0.5 x 821 x 0.36 x 0.36 x v'² = 5977.5 + 0.5 x 821 x v'² from (i)
1992.5 = 357.3 v'²
v' = 5.58 m/s
v = 0.36 x 5.58
v = 2 m/s
Rate of flow = A x v = 3.14 x 0.405 x 0.405 x 2 x 2 = 2.06 m³/s
Thus the rate of flow of volume is 2.06 m³/s.
Answer:
= 22.41rad/s
Explanation:
First, we know that:
a = 4 rad/s^2
S = 10 rev = 62.83 rad
Now we know that:
where is the final angular velocity,
the initial angular velocity, a is the angular aceleration and S the radians.
Replacing, we get:
Finally, solving for :
= 22.41rad/s