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Likurg_2 [28]
3 years ago
12

in bohr's atomic theory when electron moves from one energy level to another energy level more distant from the nucleus energy i

s emitted right?
Physics
1 answer:
Katyanochek1 [597]3 years ago
3 0
No, it is the other way around. When an electron moves from one energy level to another energy level more distant from the nucleus, it gains, not emits energy. The closer it gets to the nucleus, the more energy it emits. If it is far from the nucleus, it gains more energy. 
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Pachacha [2.7K]
D?.... I think? My guess anyway
8 0
4 years ago
A 2 ft x 2 ft x 2 ft box weighs 100 pounds, and the weight is evenly distributed. What is the magnitude of the minimum horizonta
Tems11 [23]

Answer:

Explanation:

Let the force required be F . It is applied at the top of the box . The box is likely to turn about a corner . Torque of this force about this corner

= F x 2

This torque will try to turn the box . On the other hand the weight which is acting at CM will create a torque about the same corner . This torque will try to prevent the box to turn around the corner.

This torque of weight

= 100 x 1

= 100 pound ft.

For equilibrium

Torque of F = torque of weight.

F x 2  = 100

F = 50 pounds .

7 0
4 years ago
Read 2 more answers
A 2kg body is moving a long a horizontal circular path of radius 2m with constant speed of 6m/s . what is the magnitude of the f
kondor19780726 [428]
The magnitude of force acting on the body is 36N
7 0
3 years ago
In another solar system is planet Driff, which
Fed [463]

Answer:

It is (1/5)th as much.

Explanation:

If we apply the equation

F = G*m*M / r²

where

m = mass of a man

M₀ = mass of the planet Driff

M = mass of the Earth

r₀ = radius of the planet Driff

r = radius of the Earth

G = The gravitational constant

F = The gravitational force on the Earth

F₀ = The gravitational force on the planet Driff

g = the gravitational acceleration on the surface of the earth

g₀ = the gravitational acceleration on the surface of the planet Driff

we have

F₀ = G*m*M₀ / r₀² = G*m*(5*M) / (5*r)²    

⇒  F₀ = G*m*M / (5*r²) = (1/5)*F

If

F₀ = (1/5)*F

then

W₀ = (1/5)*W   ⇒  m*g₀ = (1/5)*m*g   ⇒   g₀ = (1/5)*g

It is (1/5)th as much.

5 0
3 years ago
I'm stuck on question 3. Could anyone please explain it?
sergiy2304 [10]
Peak voltage is 2
period is 40ms
frequency = 1/period = 25Hz
5 0
3 years ago
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