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madreJ [45]
3 years ago
8

the wattage ratting of a light bulb is the power it sonsumers when it is connected accross a 120 V potential difference what is

the resistance of a 60 W light bulb
Physics
1 answer:
igor_vitrenko [27]3 years ago
8 0

Answer:

240 Ω

Explanation:

Resistance: This can be defined as the opposition to the flow of current in an electric field. The S.I unit of resistance is ohms (Ω).

The expression for resistance power and voltage is give as,

P = V²/R.......................... Equation 1

Where P = Power, V = Voltage, R = Resistance

Making R the subject of the equation,

R = V²/P.................... Equation 2

Given: V = 120 V, P = 60 W.

Substitute into equation 2

R = 120²/60

R = 240 Ω

Hence the resistance of the bulb = 240 Ω

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In a science lab, there is the adequate availability of raw materials for experiments such as chemicals while outside a science lab there isn't.

In a science lab, the availability of apparatuses and sources of heat is higher than outside a science lab.

<h3>What is science laboratory?</h3>

The science lab is a place where the experiments related to the science background are performed with the use of apparatus.

The science lab has plenty of equipment to perform the experiment whether related to physics or chemistry or biology.

Thus, source of heat and availability of raw material and equipment are the differences between the science activities carried out in a science laboratory and those carried out side of science laboratory.

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2 years ago
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How can you make the potential energy as high as possible in a magnetic field between one electromagnet and one piece of iron?
harkovskaia [24]

In step 1, to increase the potential energy, the iron will move towards the electromagnet.

In step 2, to increase the potential energy, the iron will move towards the electromagnet.

<h3>Potential energy of a system of magnetic dipole</h3>

The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

U = \mu B

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B = \frac{\mu_0 I}{2\pi r}

U = \mu\times  (\frac{\mu_0 I}{2\pi r} )

Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;

  • In step 1, to increase the potential energy, the iron will move towards the electromagnet.
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Learn more about potential energy in magnetic field here: brainly.com/question/14383738

7 0
2 years ago
____ developed the first model of the atom that showed the structure of the inside of an atom. A. Dalton B. Bohr C. Rutherford D
Elan Coil [88]

Answer:

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Explanation:

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6 0
4 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
What is the iodide ion concentration in a solution if the addition of an excess of 0.100 m pb(no3)2 to 42.9 ml of the solution p
Katen [24]
Moles Pbl2 =  0.8628 g  :  461.01 g/mol   = 0.001871

moles I = 2 x  0.001871 = 0.003742

[I-] = 0.003742/ 0.0429

= 0.0872 M
3 0
3 years ago
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