All categories

3 years ago

3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage dropped across each phase?

Physics

1 answer:

ZanzabumX [31]3 years ago

5 0

Answer:

$E_s = 277.13V$

Explanation:

Given

$Load\ Voltage = 480V$

Required

Determine the voltage dropped in each stage.

The relation between the load voltage and the voltage dropped in each stage is

$E_l = E_s * \sqrt3$

Where

$E_l = 480$

So, we have:

$480 = E_s * \sqrt3$

Solve for $E_s$

$E_s = \frac{480}{\sqrt3}$

$E_s = \frac{480}{1.73205080757}$

$E_s = 277.128129211$

$E_s = 277.13V$

<em>Hence;</em>

<em>The voltage dropped at each phase is approximately 277.13V</em>

You might be interested in

An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it

steposvetlana [31]

Answer:

μs = 0.36

Explanation:

While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.
This acceleration is produced by the centripetal force.
Now, this force is not a different type of force, is the net force acting on the riders in this direction.
Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.
In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.
When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:

$F_{frmax} = \mu_{s}* F_{n} = m * g (1)$

where μs= coefficient of static friction (our unknown)

As we have already said Fn = Fc.
The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:

$F_{n} = m* \omega^{2} * r (2)$

Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:

$\mu_{s} = \frac{g}{\omega^{2} r} (3)$

Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:

$\omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)$

Replacing g, ω and r in (3):
$\mu_{s} = \frac{g}{\omega^{2} r} = \frac{9.8m/s2}{(2.72rad/sec)^{2} *3.7 m} = 0.36 (5)$

3 0

3 years ago

A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro

prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope= $.2\frac{kg}{m}$

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as, $W_{1}=Force \times displacement$

= $\int\limits^{15}_0 {.2x} \, dx$ ..............(1)

= $.1\times 15^2$

=22.5 Nm

work done in lifting the sand is given as, $W_{2}=Force \times displacement$

$W_{2}=\int\limits^{15}_0 F \, dx$ .................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m= $\frac{20-18}{15-0}$

m=.133

Therefore,

$F=.133x+2$

Put value of F in equation 2

$W_{2}=\int\limits^{15}_0 (.133x+2) \, dx$

$W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm$

Therefore,

work done lifting the bucket (sand and rope) to the top of the building, $W=W_{1}+W_{2}$

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0

3 years ago

If we double the mass of the sun, what would happen to the gravitational force between the sun and earth?

Sphinxa [80]

It would increase
because the force is directly proportional to the Value of masses given

4 0

4 years ago

1. explain why mid-sized cars and SUVs have different rollover fatality rates.

iris [78.8K]

Answer:

1. They generally have poorer fuel efficiency and require more resources to manufacture than smaller vehicles, thus contributing more to climate change and environmental degradation. Their higher center of gravity increases their risk of rollovers.

2. Mid-sized cars and SUVs both have a fixed object fatality rate of 2.6/BVM. This is most likely because they have a similar mass and inertia. The force of the crash is likely to have about the same deceleration in both vehicles.

Explanation:

brainliest please

8 0

3 years ago

10 turns of wire are closely wound around a pencil as shown in the figure. when measured using a scale as shown, the length of t

Mila [183]

Answer:

a. The thickness of the wire is 2.5 mm.

b. The wire is 0.25 cm thick.

Explanation:

Number of turns of the wire = 10

The length of total turns = 25 mm

a. The thickness of the wire can be determined by;

thickness of the wire = $\frac{length of total turns}{number of turns}$

= $\frac{25}{10}$

= 2.5 mm

Therefore, the wire is 2.5 mm thick.

b. To determine the thickness of the wire in centimetre;

10 mm = 1 cm

So that,

2.5 mm = x

x = $\frac{2.5}{10}$

= 0.25 cm

The wire is 0.25 cm thick.

8 0

3 years ago