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mote1985 [20]
3 years ago
12

What year did niels bohr discover the atomic theory​

Physics
2 answers:
Ludmilka [50]3 years ago
8 0

Answer:

1913

Explanation:

<u>A hypothesis of the hydrogen atom was developed by Niels Bohr in 1913, based on the quantum theory</u>, which states that certain physical quantities can only have discrete values. Electronic particles travel about a nucleus, but only in specified orbits. If electrons jump to a lower-energy orbit, the energy difference between the two orbits is emitted as radiation.

Hope this helps explain what the answer to your question is and why it is that answer.

Dimas [21]3 years ago
7 0

Answer:

1913

Explanation:

In 1913, Niels Bohr proposed a theory for the hydrogen atom, based on quantum theory that some physical quantities only take discrete values. Electrons move around a nucleus, but only in prescribed orbits, and If electrons jump to a lower-energy orbit, the difference is sent out as radiation.

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alukav5142 [94]
Without an external agent doing work, heat will always flow from a hotter to a cooler object. Two objects of different temperature always interact. There are three different ways for heat to flow from one object to another. They are conduction, convection, and radiation.
5 0
3 years ago
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A swing has a period of 10 seconds. What is its frequency ?
kow [346]

Frequency = 1 / (period)

Frequency = 1 / (10 seconds) = (1/10) ( / second) = 0.1 per second = <em>0.1 Hz</em>.


4 0
3 years ago
Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
netineya [11]

Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)

Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg

Δu=1300kJ/kg  

3 0
3 years ago
How to do this, i'm completely lost
vaieri [72.5K]
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g 
t2 = 7 * 49g 
t3 = 3.5 * 24g 

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°)) 

T = 1501,6 N

8 0
3 years ago
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scoray [572]

Answer:

heck ya its so fire

Explanation:

3 0
3 years ago
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