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2 years ago

A t-rated switch may be used to (its/a) ______ current capacity when controlling an incandescent lighting load.

Physics

1 answer:

jok3333 [9.3K]2 years ago

6 0

<u>Rated </u>

A t-rated switch may be used to (its/a) <u>Rated</u> current capacity current capacity when controlling an incandescent lighting load.

<h3>What is "current rating"?</h3>

The greatest current that a fuse is rated to carry for an infinite duration without significantly degrading the fuse element is known as the current rating.
There is also a large selection of power switching transistors that have voltage ratings well over 1000V and current ratings up to several hundred amps.

<h3>What does the term "rated current" mean?</h3>

When an electrical device receives its rated voltage and outputs its rated power, it flows at its rated current.
So, when a device is designed for a certain amperage, that amperage is referred to as the equipment's rated current.

<h3>What does electrical "rated" mean?</h3>

An electrical appliance's rating reveals the voltage range at which it is intended to operate as well as the current consumption at that range.
These numbers are typically shown on a rating plate that is fastened to the device, such as 230 volts, 3 amps.

To learn more about current capacity visit:

brainly.com/question/27987428

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A 15.0kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0 degrees above the horizontal. The bl

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Answer:

$W=70 * 5cos20 = 328.89 J$

$W_n = 0$

$W_g=0$

$W_f= -184.59J$

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Explanation:

From the question we are told that

Weight of block =15.0kg

Force acting on the block = 70.0N

At an angle of 20 degree

Displacement of block is 5m

Coefficient of kinetic friction 0.3

b) Generally work done by force is give by $W=fdcos \theta$

therefore

$W=70 * 5cos20 = 328.89 J$

c) there is no work done by the normal force in this scenario because

normal force in this case is perpendicular to the displacement of the motion

$W_n = 0$

d) The displacement in the vertical direction is 0

Therefore the gravitational work done is 0 $W_g=0$

e)Generally in finding work done by friction we first find frictional force

Mathematically the equation for frictional force is given $f = \alpha N$

Given that

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$N=123 N$

$f=0.3* 123.06 = 36.92N$

Mathematically solving to get work done by frictional force $W_f$

$W_f= -fd\\W_f = -36.92 * 5$

$W_f= -184.59J$

the frictional force work done is $W_f= -184.59J$

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