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2 years ago

The source of heat for a (an) _____ system is electricity

Physics

1 answer:

marshall27 [118]2 years ago

3 0

Electric Heating systems use electricity as a source of heat.

<u>Explanation: </u>

Each heater consists of an electric resistor, which the heating element in it. The process involved is Joule Heating. It is the phenomenon in which, passage of electric current produces heat. Heat is generated when the electrons from the current and the positive ions of the conductor start interaction.

The presence of voltage difference makes an electric fields, which in turn accelerates the charges present in the conductors. When they collide with ions of conductor, heat gets generated, and emitted out, which we use for heating purposes.

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Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

$a(t)=(5-10t)\ m/s^2$

(a) Since, $a=\dfrac{dv}{dt}$

v = velocity

$dv=a.dt$

$v=\int(a.dt)$

$v=\int(5-10t)(dt)$

$v=5t-\dfrac{10t^2}{2}=5t-5t^2$

(b) $v=\dfrac{dx}{dt}$

x = position

$x=\int v.dt$

$x=\int (5t-5t^2)dt$

$x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3$

(c) Velocity function is given by :

$v=5t-5t^2$

$5t-5t^2=0$

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

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2 years ago

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Answer:

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3 years ago

A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie

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Explanation:

It is given that,

Area of nickel wire, $A=7.4\ cm^2=7.4\times 10^{-4}\ m^2$

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, $B_1=0.5\ T$

Final magnetic field, $B_2=3\ T$

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{B_2-B_1}{t}$

$\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}$

$\epsilon=1.65\times 10^{-3}\ V$

Induced current in the loop of wire is given by :

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{1.65\times 10^{-3}}{2.4}$

$I=6.87\times 10^{-4}\ A$

So, the induced current in the loop of wire over this time is $6.87\times 10^{-4}\ A$ . Hence, this is the required solution.

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3 years ago

A ray of yellow light (f = 5.09 × 1014hz) travels at a speed of 2.04 × 108meters per second in

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Velocity = fλ

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4.007*10⁻⁷ m = λ

The wavelength of the yellow light = 4.007*10⁻⁷ m

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3 years ago

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Answer:

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