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3 years ago

15

In a certain balanced three phase system each line current is a 5a and each line voltage is 220v . What is the approximate real

power if the power factor is 0.7

1 answer:

lara [203]3 years ago

4 0

Answer:

1,334

Explanation:

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Turbine blades mounted to a rotating disc in a gas turbine engine are exposed to a gas stream that is at [infinity] = 1100°C and

Nataly_w [17]

Answer:

<u><em>note:</em></u>

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3 years ago

______________ help protect the lower legs and feet from heat hazards like molten metal and welding sparks. A) Safety shoesB) Le

vodka [1.7K]

Answer:

It's leggings. They help protect the lower legs and feet from heat hazards like molten metal and welding sparks.

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3 years ago

Lets Try This: study the pictures. Describe what you see and think about it. write your answer on a sheet of paper. home room

Yuliya22 [10]

Answer: I see three children cleaning the lawn while one of them are raking the leaves and one is holding a dust pan. The other child is holding a bucket. On the other picture, i see a young boy watering plants.

BTW: these pictures are not very clear so answers may vary.

Explanation:

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2 years ago

A rigid, sealed tank initially contains 2000 kg of water at 30 °C and atmospheric pressure. Determine: a) the volume of the tank

Bad White [126]

Given:

mass of water, m = 2000 kg

temperature, T = $30^{\circ}C$ = 303 K

extacted mass of water = 100 kg

Atmospheric pressure, P = 101.325 kPa

Solution:

a) Using Ideal gas equation:

PV = m $\bar{R}$ T (1)

where,

V = volume

m = mass of water

P = atmospheric pressure

$\bar{R} = \frac{R}{M}$

R= Rydberg's constant = 8.314 KJ/K

M = molar mass of water = 18 g/ mol

Now, using eqn (1):

$V = \frac{m\bar{R}T}{P}$

$V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}$

$V = 2762.44 m^{3}$

Therefore, the volume of the tank is $V = 2762.44 m^{3}$

b) After extracting 100 kg of water, amount of water left, m' = m - 100

m' = 2000 - 100 = 1900 kg

The remaining water reaches thermal equilibrium with surrounding temperature at T' = $30^{\circ}C$ = 303 K

At equilibrium, volume remain same

So,

P'V = m' $\bar{R}$ T'

$P' = \frac{1900\times \frac{8.314}{18}\times 303}{2762.44}$

Therefore, the final pressure is P' = 96.258 kPa

4 0

3 years ago

simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The

Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

$q(x)=\frac{x}{L} q_{0}$

σ = 120 MPa = 120*10⁶ Pa

$L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\$

We can see the pic shown in order to understand the question.

We apply

∑MB = 0 (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6 (↑)

Then, we apply

$v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x$

Then, we can get the maximum bending moment as follows

$M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m$

then we get

$M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}$

We get the inertia as follows

$I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}$

We use the formula

σ = M*y/I

⇒ M = σ*I/y

where

$y=\frac{h}{2} =\frac{0.3m}{2}=0.15m$

If M = Mmax, we have

$(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}$

8 0

4 years ago