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anastassius [24]
3 years ago
5

3/25 of the current delivered to an induction motor is wasted as reactive current.

Mathematics
2 answers:
r-ruslan [8.4K]3 years ago
8 0
Answer: 0.12

Explanation:

3/25 = 0.12
sp2606 [1]3 years ago
3 0

Answer:

0.12

Step-by-step explanation:

3/25= 12/100

12/100=0.12

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Answer:

y is equal to 4.

Step-by-step explanation:

To find this, cross multiply and then divide.

10*2 = y*5

20 = 5y

4 = y

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Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat
nirvana33 [79]

Answer:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

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And replacing we got:

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And that represent 0.127 \frac{cm}{min}

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}

And solving for \frac{dh}{dt} we got:

\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}

We need to convert the rate given into m^3/min and we got:

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

5 0
3 years ago
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