The flow velocity measured as in/min at the meter is 5924.13 in/min
<h3 /><h3>Volumetric flow rate</h3>
We know that the volume flow rate Q = Av where
- A = cross-sectional area of pipe and
- v = flow velocity
Now, Q = 18 gal/min
Converting this to in³/min, we have
Q = 18 gal/min = 18 × 1 gal/min = 18 × 231 in³/min = 4158 in³/min
A = πd²/4 where d = diameter of pipe = 1.0 in.
<h3 /><h3>Flow velocity, v</h3>
Since Q = Av, making v subject of the formula, we have
v = Q/A
v = 4Q/πd²
Substituting the values of the variables into the equation, we have
v = 4Q/πd²
v = 4 × 4158 in³/min ÷ π × (1.0 in)²
v = 16632 in/min ÷ π
v = 5924.13 in/min
So, the flow velocity measured as in/min at the meter is 5924.13 in/min
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Answer:
f(2)=1/9
Step-by-step explanation:
f(x)=(1/3)^x
f(2)=(1/3)^(2)
f(2)=1/9
What are you trying to ask ??
Answer:
4) 3.
Step-by-step explanation:
According to the graph, f(x) has been shifted two units to the right and 3 units up. k is the number of units f(x) has been shifted vertically. So, the value of K is 4) 3.
Hope this helps!
Answer:
cos 2Ф = - 161/289 , tan 2Ф = - 240/161
Step-by-step explanation:
* Lets explain how to solve the problem
∵ cos Ф = - 8/17
∵ Ф lies in the 3rd quadrant
- In the 3rd quadrant sin and cos are negative values, but tan is
a positive value
∵ sin²Ф + cos²Ф = 1
∴ sin²Ф + (-8/17)² = 1
∴ sin²Ф + 64/289 = 1
- Subtract 64/289 from both sides
∴ sin²Ф = 225/289 ⇒ take √ for both sides
∴ sin Ф = ± 15/17
∵ Ф lies in the 3rd quadrant
∴ sin Ф = -15/17
∵ cos 2Ф = 2cos²Ф - 1 ⇒ the rule of the double angle
∵ cos Ф = - 8/17
∴ cos 2Ф = 2(-8/17)² - 1 = (128/289) - 1 = - 161/289
* cos 2Ф = - 161/289
∵ tan 2Ф = sin 2Ф/cos 2Ф
∵ sin 2Ф = 2 sin Ф × cos Ф
∵ sin Ф = - 15/17 and cos Ф = - 8/17
∴ sin 2Ф = 2 × (-15/17) × (-8/17) = 240/289
∵ cos 2Ф = - 161/289
∴ tan 2Ф = (240/289)/(-161/289) = - 240/161
* tan 2Ф = - 240/161
