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sveta [45]
3 years ago
9

Which scientist was the first to conclude through experimentation that atoms have positive charges in their nuclei?

Physics
1 answer:
Pavel [41]3 years ago
8 0
<h2>Answer: Ernest Rutherford </h2>

Ernest Rutherford was a British physicist and chemist of New Zealand origin, who conducted a series of experiments together with Hans Geiger and Ernest Marsden; where the result led him to propose a new atomic model.

It should be noted that at that time, the "accepted" atomic model was Thomson's raisin pudding atomic model<u> </u><em><u>(electrons with negative charge immersed an the atom of positive charge that counteracted the negative charge of the electrons, like raisings embedded in a pudding)</u></em>, who discovered the electron and formerly was a professor of Rutherford.  

Now, the experiment conducted under the direction of Ruherford at the laboratories of the University of Manchester during the year 1911; was for the purpose of <u>corroborating Thomson's atomic model</u>. To achieve this, a thin metal sheet was bombarded with alpha particles (nuclei of helium gas).

The idea was that these alpha particles, having positive electric charge, were attracted by the atom's negative charges and repelled by the positive charges, and it was expected that they would pass through the thin sheet without hardly deviating. Then, to observe the crash site of the particle, a phosphorescent screen was placed behind and on the sides of the metal sheet.

For according to Thomson's atomic model the positive and negative charges were evenly distributed, the sphere (the atom) had to be electrically neutral, and <u>the alpha particles would pass through the sheet without deviating. </u>

However, the results were surprising:

As expected, most of the particles went through the sheet without deviating.

<h2>But some suffered large deviations and, most importantly, <u>a small number of particles bounced backwards</u>. </h2>

That is:

<h2>The alpha particle beam was scattered (repelled) when it hit the thin metal sheet. </h2>

These facts could not be explained by Thomson's atomic model, so Rutherford developed another, suggesting that:

<h2><em>There is a concentration of charge in the center of the atom (which was later called nucleus) surrounded by electrons. </em></h2>

This new model could explain the proven fact in his experiments that some particles bounced in the direction opposite to the incident particles, because the electrical charge of this nucleus was positive, equal to the electrical charge of the alpha particles.

This is how Rutherford proposed a new atomic model and discovered the existence of the nucleus. However, this was not the definitive model, because on 1913 it was replaced by Bohr's.

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A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simp
masha68 [24]

To solve this problem we must rely on the equations of the simple harmonic movement that define the period as a function of length and gravity as

T = 2\pi \sqrt{\frac{l}{g}}

Where

l = Length

g = Gravity

Re-arrange to find L,

L = g (\frac{T}{2\pi})^2

Our values are given as

g = 9.81m/s

T = 10.1s

Replacing,

L = g (\frac{T}{2\pi})^2

L = (9.81) (\frac{10.1}{2\pi})^2

L = 25.348m

Therefore the height would be 25.348m

5 0
3 years ago
A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformati
Slav-nsk [51]

The question is incomplete. The complete question is :

A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformation cycle. At a time, t = 0, a tensile stress of 20 MPa is applied instantaneously and maintained for 100 s. The stress is then removed at a rate of 0.2 MPa s−1 until the polymer is unloaded. If the creep compliance of the material is given by:

J(t) = Jo (1 - exp (-t/to))

Where,

Jo= 3m^2/ GPA

to= 200s

Determine

a) the strain after 100's (before stress is reversed)

b) the residual strain when stress falls to zero.

Answer:

a)-60GPA

b) 0

Explanation:

Given t= 0,

σ = 20Mpa

Change in σ= 0.2Mpas^-1

For creep compliance material,

J(t) = Jo (1 - exp (-t/to))

J(t) = 3 (1 - exp (-0/100))= 3m^2/Gpa

a) t= 100s

E(t)= ΔσJ (t - Jo)

= 0.2 × 3 ( 100 - 200 )

= 0.6 (-100)

= - 60 GPA

Residual strain, σ= 0

E(t)= Jσ (Jo) ∫t (t - Jo) dt

3 × 0 × 200 ∫t (t - Jo) dt

E(t) = 0

5 0
3 years ago
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m
marusya05 [52]

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

4 0
3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
2 years ago
A1.200 kg car is sliding down an icy
Triss [41]

Answer:

1585.67N

Explanation:

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