The primary reason a light bulb emits light is due to

Physics

2 answers:

RideAnS [48]3 years ago

6 0

Answer: Short answer is resistance.

Mashutka [201]3 years ago

5 0

The primary reason a light bulb emits light is due to the heating of the resistance in the filament of the light bulb. In fact, the power dissipated in a resistor is given by
 $P=I^2 R$
where I is the current and R the resistance. The larger the resistance or the current in the resistor, the larger the power dissipated. Due to this dissipation of power, the temperature of the filament becomes very high, and the resistance becomes incandescent, emitting light.

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Y=mx+b if y=8.18 m=1.31 b=17.2 then find x

Dafna1 [17]

Plug in the corresponding values into y = mx + b

8.18 in for y

1.31 in for m

17.2 in for b

8.18 = 1.31x + 17.2

Now bring 17.2 to the left side by subtracting 17.2 to both sides (what you do on one side you must do to the other). Since 17.2 is being added on the right side, subtraction (the opposite of addition) will cancel it out (make it zero) from the right side and bring it over to the left side.

8.18 - 17.2 = 1.31x

-9.02 = 1.31x

Then divide 1.31 to both sides to isolate x. Since 1.31 is being multiplied by x, division (the opposite of multiplication) will cancel 1.31 out (in this case it will make 1.31 one) from the right side and bring it over to the left side.

-9.02/1.31 = 1.31x/1.31

x ≈ -6.8855

x is roughly -6.89

Hope this helped!

~Just a girl in love with Shawn Mendes

8 0

3 years ago

Read 2 more answers

The first law of thermodymaic stae the 5ktiopheuithkjhuiguihaoitg

pentagon [3]

I'm assuming you want the first law of thermodynamics.

The First Law of Thermodynamics states that heat is a form of energy and cannot be created or destroyed. It can, however, be transferred from one location to another and can be converted into other forms of energy.

3 0

3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by