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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 statements accurately describe Dmitri Mendeleev’s contributions to the development of the periodic table? Check all that a
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He produced the first orderly arrangement of known elements, he used patterns to predict undiscovered elements
6 0
3 years ago
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Why is the motion of an athlete moving along the circular path with Constant speed considered to be an accelerated motion?
Strike441 [17]

The speed is changing its direction all the time. There is an acceleration which changes the direction of the speed – that is called centripetal acceleration. Only uniform linear motions are considered to have no acceleration.

This is the general formula for acceleration

a = dv/dt

When calculating dv, you should keep in mind the change in the velocity vector’s direction. You can easily see in a graph that with dt tending to 0 (so the length of the arc covered is also tending to 0), the difference between vectors Vf and V0 has a direction which is perpendicular to velocity (the shorter the arc, the closest the angle is to 90 degrees).

There is a formula (which can be deducted from the previous formula) which allows you to calculate the acceleration:

a = v^2/r

Let’s talk about the units:

v is in m/s

r is in m

so v^2/r

is in (m/s)^2/m = (m^2/s^2)/m = m/s^2

which is the same unit as dv/dt:

dv/dt = (m/s)/s= m/s^2

5 0
3 years ago
Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
3 years ago
15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remai
alexandr1967 [171]

The alpha particle is emitted at 4235 m/s

Explanation:

We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:

p_i = p_f\\ Mu=m_1 v_1 + m_2 v_2 =  

where:  

M =222u is the mass of the original nucleus

v=420 m/s is the initial velocity of the nucleus

m_1 = 4 u is the mass of the alpha particle

v_1 is the final velocity of the alpha particle

m_2 = 222u-4u = 218 u is the mass of the daughter nucleus

v_2 = 350 m/s is the final velocity of the nucleus

Solving for v_1, we  find the final velocity of the alpha particle:

v_1 = \frac{Mu-m_2 v_2}{m_1}=\frac{(222)(420)-(218)(350)}{4}=4235 m/s

Learn more about momentum:

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3 years ago
What human practices are destroying the rain forest?
marissa [1.9K]
Deforestation, the chopping off of the trees that can take thousands or even millions of years to grow again<span />
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