How many inner shell electrons are there in Bohr model

Physics

1 answer:

Paha777 [63]3 years ago

4 0

Answer:

10 is the correct answer

Explanation:

It is ten because the electrons that encircled the two orbit before the third orbit if the bohr model is 10. In bohr model, the electrons encircled the nucleus of an atom called orbit.

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Two bricks are released at the same time from the same point ten feet above the ground. One of the bricks is falling straight do

natita [175]

Both bricks will hit the ground at the same time.

Falling vertically is always accelerating at 9.8 m/s² because of gravity.
Nothing that's happening horizontally has any effect on that.
The brick that happens to have some horizontal motion will
probably hit the ground way over there, but that will still be
at the same TIME as this one.

This is a perfect place to remind you of the old unbelievable story,
which I'll bet you heard before:

If you fire a bullet horizontally from a gun, and at the exact same
moment you DROP another bullet out of your hand next to the gun,
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Horizontal speed has no effect on vertical behavior.

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3 years ago

Air "breaks down" when the electric field strength reaches 3×106n/c, causing a spark. a parallel-plate capacitor is made from tw

aalyn [17]

Ok, I think this is right but I am not sure:
Q = ϵ
0AE
A= π π
r^2

=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
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3 years ago

Find the instantaneous velocity at 1 s . can anyone help with c-h!!!

Lelu [443]

Rise over run at 1 second
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7 0

3 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$