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Tasya [4]
3 years ago
12

The Bohr model was determined ultimately to be flawed. Knowing that "opposites attract (and 'likes' repel)," can you think of an

y flaws in the Bohr model that disobey these statements?
Physics
1 answer:
Sloan [31]3 years ago
6 0

Answer:

Yes, there are 2 flaws

Explanation:

Electromagnetics say that, given two particles with charge, they will be attracted if they have opposite charge or repelled if they have identical charge.

Then, we can find two flaws in the Bohr model. The first one is that the electrons move in energy layers (or just layers to get the idea) far from the proton compared with the distance between the electrons themselves. So, if the model was right, how can it be that they don't repel each other? With the same logic, the protons don't repel each other either even though they are all together in the nucleus.

The second flaw, related to what we've just said, is that the electrons can move from one layer to another, but they will always stay at a minimum distance to the proton. How can it be so if it is known that they attract because of their charge signs?

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Which device is used to measure interior diameter of a water pipe and in which units are the callibrations​
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We can use a vernier calliper

3 0
2 years ago
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Prove that the unit of area derived quantities are derived units​
frozen [14]

answer: derived physical quantities are those quantities that are obtained from the basic physical quantities by multiplication or division and area is one of them

8 0
3 years ago
Add the vectors:
Anettt [7]

Vector 1 has components

x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m

y_1=(10\,\mathrm m)\sin20^\circ\approx3.42\,\mathrm m

and vector 2 has

x_2=(10\,\mathrm m)\cos80^\circ\approx1.74\,\mathrm m

y_2=(10\,\mathrm m)\sin80^\circ\approx9.85\,\mathrm m

Add these vectors to get the resultant, which has components

x_{\rm total}\approx11.133\,\mathrm m

y_{\rm total}\approx13.268\,\mathrm m

The magnitude of the resultant is

\sqrt{{x_{\rm total}}^2+{y_{\rm total}}^2}\approx17.321\,\mathrm m

with direction \theta such that

\tan\theta=\dfrac{y_{\rm total}}{x_{\rm total}}\implies\theta\approx50^\circ

or about 50º N of E.

8 0
3 years ago
Find the final velocity of a car that accelerates at +2 m/s2 for +4m from an
Stolb23 [73]

Answer:

Explanation:

according to third equation of motion

2as=vf²-vi²

vf²=2as+vi²

vf=√2as+vi²

vf=√2as+vi

vf=√2*2*4+3

vf=√16+3

vf=4+3=7

so final velocity is 7 m/s

5 0
3 years ago
6. A .25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. a. What is the final
Alenkasestr [34]

Answer:

(a) the final velocity of the combined mass is 9.43 m/s

(b) the decrease in kinetic energy during the collision is 386.1 J

Explanation:

Given;

mass of arrow, m₁ = 25 kg

initial velocity of arrow, u₁ = 12 m/s

mass of target, m₂ = 6.8 kg

initial velocity of the target, u₂ = 0

Part (a)

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final velocity of the combined mass

25 x 12 + 0 = v(25 + 6.8)

300 = v(31.8)

v = 300/31.8

v = 9.43 m/s

Part(b)

Kinetic Energy, K.E = ¹/₂mv²

Initial kinetic energy =  ¹/₂m₁u₁² + ¹/₂m₂u₂²  = ¹/₂ x 25 x (12)² + 0 = 1800 J

Final kinetic energy = ¹/₂m₁v² + ¹/₂m₂v² = ¹/₂v²(m₁ + m₂)

                                                               = ¹/₂ x (9.43)²(25+6.8)

                                                               = 1413.91 J

Decrease in kinetic energy = Initial K.E - Final K.E

Decrease in kinetic energy = 1800J - 1413.91 J = 386.1 J

                               

4 0
3 years ago
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