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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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What is the force of a 13kg ball that has been dropped and has fallen for 1 second?
ankoles [38]

Answer:

127.4 newtons

Explanation:

Assuming g = 9.8:

F = ma = 13(9.8) = 127.4 N

6 0
3 years ago
A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0
4 years ago
A) In the figure below, a cylinder is compressed by means of a wedge against an elastic constant spring = 12 /. If = 500 , deter
Radda [10]

Explanation:

A) Draw free body diagrams of both blocks.

Force P is pushing right on block A, which will cause it to move right along the incline.  Therefore, friction forces will oppose the motion and point to the left.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force N pushing up and left 10° from the vertical,

Friction force Nμ pushing down and left 10° from the horizontal,

Reaction force Fab pushing down,

and friction force Fab μ pushing left.

There are 2 forces acting on block B:

Reaction force Fab pushing up,

And elastic force kx pushing down.

(There are also horizontal forces on B, but I am ignoring them.)

Sum of forces on A in the x direction:

∑F = ma

P − N sin 10° − Nμ cos 10° − Fab μ = 0

Solve for N:

P − Fab μ = N sin 10° + Nμ cos 10°

P − Fab μ = N (sin 10° + μ cos 10°)

N = (P − Fab μ) / (sin 10° + μ cos 10°)

Sum of forces on A in the y direction:

N cos 10° − Nμ sin 10° − Fab = 0

Solve for N:

N cos 10° − Nμ sin 10° = Fab

N (cos 10° − μ sin 10°) = Fab

N = Fab / (cos 10° − μ sin 10°)

Set the expressions equal:

(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)

Cross multiply:

(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)

Distribute and solve for Fab:

P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)

P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)

Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)

Sum of forces on B in the y direction:

∑F = ma

Fab − kx = 0

kx = Fab

x = Fab / k

x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))

Plug in values and solve.

x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))

x = 0.0408 m

x = 4.08 cm

B) Draw free body diagrams of both blocks.

Force P is pushing block A to the right relative to the ground C, so friction force points to the left.

Block A moves right relative to block B, so friction force on A will point left.  Block B moves left relative to block A, so friction force on B will point right (opposite and equal).

Block B moves up relative to the wall D, so friction force on B will point down.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force Fc pushing up,

Friction force Fc μ₁ pushing left,

Reaction force Fab pushing down and left 15° from the vertical,

and friction force Fab μ₂ pushing up and left 15° from the horizontal.

There are 5 forces acting on block B:

Weight force 750 n pushing down,

Normal force Fd pushing left,

Friction force Fd μ₁ pushing down,

Reaction force Fab pushing up and right 15° from the vertical,

and friction force Fab μ₂ pushing down and right 15° from the horizontal.

Sum of forces on B in the x direction:

∑F = ma

Fab μ₂ cos 15° + Fab sin 10° − Fd = 0

Fd = Fab μ₂ cos 15° + Fab sin 15°

Sum of forces on B in the y direction:

∑F = ma

-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0

Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Substitute:

(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750

Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)

Sum of forces on A in the y direction:

∑F = ma

Fc + Fab μ₂ sin 15° − Fab cos 15° = 0

Fc = Fab cos 15° − Fab μ₂ sin 15°

Sum of forces on A in the x direction:

∑F = ma

P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0

P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁

Substitute:

P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁

P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°

P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)

First, find Fab using the given values.

Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)

Fab = 1151.9 N

Now, find P.

P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)

P = 1095.4 N

6 0
3 years ago
The properties of ductility, malleability, ability to conduct heat and electricity are characteristics of what type of material?
laiz [17]
These are  the characterstics of metal.


                 hope it helps u........
8 0
4 years ago
Read 2 more answers
A particle's position along the x-axis is described by. x(t)= At+Bt^2where t is In seconds: x is in meters: and the constants A
DanielleElmas [232]

Answer

given,

position of particle

x(t)= A t + B t²

A = -3.5 m/s

B = 3.9 m/s²

t = 3 s

a)  x(t)= -3.5 t + 3.9 t²

   velocity of the particle is equal to the differentiation of position w.r.t. time.

\dfrac{dx}{dt}=\dfrac{d}{dt}(-3.5t + 3.9t^2)

v= -3.5 + 7.8 t------(1)

velocity of the particle at t = 3 s

  v = -3.5 + 7.8 x 3

 v = 19.9 m/s

b) velocity of the particle at origin

  time at which particle is at origin

  x(t)= -3.5 t + 3.9 t²

   0 = t (-3.5  + 3.9 t )

   t = 0, t=\dfrac{3.5}{3.9}

   t = 0 , 0.897 s

speed of the particle at t = 0.897 s

from equation (1)

 v = -3.9 + 7.8 t

 v = -3.9 + 7.8 x 0.897

  v = 3.1 m/s

6 0
4 years ago
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