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3 years ago

What is an electric load ?write with example

Physics

1 answer:

Ganezh [65]3 years ago

4 0

Explanation:

an electrical load is the part of an electrical circuit in which current is transformed into something useful. examples include a lightbulb, a resistor and a motor. a load converts electricity into heat, light or motion. put another way, the part of a circuit that connects to a well-defined output terminal is considered an electrical load.

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he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?

Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

$\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}$

With the following condition :

$\sf{\omega}$ = angular frequency (rad/s)
$\sf{\theta}$ = change of angle value (rad)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

$\sf{\theta}$ = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
t = interval of the time = 54.9 s.

What was asked :

$\sf{\omega}$ = angular frequency = ... rad/s

Step by step :

$\sf{\omega = \frac{\theta}{t}}$

$\sf{\omega = \frac{2,000 \pi}{54.9}}$

$\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}$

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0

2 years ago

A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s

ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0

3 years ago

A key falls from a bridge that is 32 m above the water. It falls directly into a model boat, moving with constant velocity, that

FinnZ [79.3K]

Answer:

Speed of the boat, v = 4.31 m/s

Explanation:

Given that,

Height of the bridge, h = 32 m

The model boat is 11 m from the point of impact when the key was released, d = 11 m

Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$t=\sqrt{\dfrac{2s}{g}}$

$t=\sqrt{\dfrac{2\times 32}{9.8}}$

t = 2.55 seconds

Let v is the speed of the boat. It can be calculated as :

$v=\dfrac{d}{t}$

$v=\dfrac{11\ m}{2.55\ s}$

v = 4.31 m/s

So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.

3 0

3 years ago

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm