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1 year ago

How many sub levels does the third energy level have?

Physics

1 answer:

EastWind [94]1 year ago

3 0

The energy levels are also called principal levels. The number of sublevels in the third energy level is three.

<h3>What is energy level?</h3>

The Energy levels are at a fixed distance from the nucleus of an atom, where electrons are arranged in shells.

In the third energy shell, electrons are present. The third principal energy level has three sublevels, s, p and d.

The number of sublevels in the third energy level is three.

Learn more about energy levels.

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Helpp pls

Kipish [7]

Answer:

The intensity of the electric field is

$|E|=10654.37 \:N/C$

Explanation:

The electric field equation is given by:

$|E|=k\frac{q}{d^{2}}$

Where:

k is the Coulomb constant
q is the charge at 0.4100 m from the balloon
d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

$|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}$

$|E|=10654.37 \:N/C$

I hope it helps you!

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3 years ago

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2 years ago

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3 years ago

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What is the reaction force if a girl pulls on a cow?

Papessa [141]

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3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

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The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).