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viktelen [127]
3 years ago
11

Why has the model of the atom changed over time?

Physics
1 answer:
Makovka662 [10]3 years ago
7 0
<span>A.
The model of the atom evolved as scientists have made new discoveries.</span>
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How much current must be applied across a 60 Ω light bulb filament in order for it to consume 55 W of power? Unserious answers w
sergij07 [2.7K]

Answer: current I = 0.96 Ampere

Explanation:

Given that the

Resistance R = 60 Ω 

Power = 55 W

Power is the product of current and voltage. That is

P = IV ...... (1)

But voltage V = IR. From ohms law.

Substitutes V in equation (1) power is now

P = I^2R

Substitute the above parameters into the formula to get current I

55 = 60 × I^2

Make I^2 the subject of formula

I^2 = 55/60

I^2 = 0.92

I = sqr(0.92)

I = 0.957 A

Therefore, 0.96 A current must be applied.

4 0
3 years ago
Pls helpppppp i’ll give brainliest
Tema [17]

Answer:

D

Explanation:

The answer is Niels Bohr's planetary model, the difference between this model and all of the other models is that the Bohr's PM Is more of layers of

Nucleus - Protons and Neutrons

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Electron Orbital - Period 2 Elements

8 electrons

Electron Orbital - Period 3 Elements

8 electrons

If that made sense-

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3 years ago
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What is the outermost layer of the sun? photosphere corona core radiative zone chromosphere convective zone
murzikaleks [220]
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Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge
NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is V=\frac{1}{4\pi\epsilon} \frac{q}{d}

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

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3 years ago
Where should an object be placed in front of a concave mirror’s principal axis to form an image that is real, inverted, larger t
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<em> I believe the answer is letter C.</em>
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