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Vikki [24]
1 year ago
15

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

Physics
1 answer:
jok3333 [9.3K]1 year 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

#SPJ4

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What is an advantageous disadvantage of generating electricity using a hydroelectric power station rather than wind turbines ?
Andrei [34K]

The advantages of generating electricity using a hydroelectric power station rather than wind turbines is that it's not polluting anything since it's regular water. It also can be used over and over again.

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It's better than wind turbines because wind turbines require a lot of wind to turn and on days without wind, very little energy can be produced.

7 0
3 years ago
What would happen to the moon if earth stopped exerting the force of gravity on it?
Mila [183]
There are two equal forces of gravity between the Earth and the Moon.
One force pulls the Moon toward the Earth.
The other force pulls the Earth toward the Moon.

If only this gravity suddenly switched off, then the moon would
continue to orbit the Sun, very much as it does now.

If ALL gravity suddenly switched off, then . . .

-- the Moon would stop orbiting the Earth and would sail away, in
a straight line and at the speed it had when gravity disappeared;

-- the Earth would stop orbiting the Sun and would sail away, in
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-- all the gases surrounding the Earth ... which we call "air" ... would
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-- the Sun would completely fall apart, expand into a giant cloud of gas,
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6 0
3 years ago
Read 2 more answers
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
NeTakaya

Answer:

Explanation:

General equation of the electromagnetic wave:

E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]

where

\phi = Phase angle, 0

c = speed of the electromagnetic wave, 3 × 10⁸

\lambda = wavelength of electromagnetic wave, 698 × 10⁻⁹m

E₀ = 3.5V/m

Electric field equation

E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]

Magnetic field Equation

B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]

Where B₀= E₀/c

B_0 = \frac{E_0}{c} = \frac{3.5}{3\times10^8}=1.2 \times 10^{-8}T

B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]

6 0
3 years ago
A 0.260 m radius, 525 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a
Sophie [7]

Answer:

B = 0.37T

Explanation:

In order to calculate the needed magnitude of the magnetic force you use the following formula, which calculate the induced emf of the solenoid when there is a change in the magnetic flux:

emf=-N\frac{\Delta \Phi_B}{\Delta t}=-N\frac{\Delta (BAcos\theta)}{\Delta t}       (1)

emf: induced voltage in the solenoid = 10,000V

N: turns of the solenoid = 525

ФB: magnetic flux

B: magnitude of the magnetic field = ?

A: cross-sectional area of the solenoid = π*r^2

r: radius of the cross-sectional area = 0.260m

Δt: interval time of the change of the magnetic flux = 4.17ms = 4.17*10^-3s

First, you have the magnetic field direction perpendicular to the plane of the solenoid, after, the angle between them is 90°  (quarter of a revolution)

In the equation (1) the only parameter that changes on time is the angle, then, you can solve for B from the equation (1):

emf=-NBA\frac{cos(90\°)-cos(0\°)}{\Delta t}=\frac{NBA}{\Delta t}\\\\B=\frac{(\Delta t)(emf)}{NA}=\frac{(\Delta t)(emf)}{N(\pi r^2)}\\\\

Finally, you replace the values of the parameters to calculate B:

B=\frac{(4.17*10^{-3}s)(10000V)}{(525)(\pi(0.260m)^2)}=0.37T

The strength of the magnetic field is 0.37T

7 0
3 years ago
The velocity of a 150 kg cart changes from 6.0 m/s to 14.0 m/s. What is the magnitude of the impulse that acted on it?
aleksandrvk [35]
M = 150 kg.

Final velocity, v = 14 m/s

Initial Velocity, u = 6 m/s

Impulse =  m(v - u)
 
             = 150*(14 - 6)
           
             =  150*8 = 1200 kgm/s  or 1200 Ns 
8 0
3 years ago
Read 2 more answers
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