All categories

3 years ago

Why did Thomson's results from experimenting with cathode rays a big change in scientific thought about atoms?

1 answer:

3 0

Thomson's results from experiments with cathode ray tubes produced a big change in scientific thought about atoms because <u>his results gave the first evidence that atoms were made of smaller particles.</u>

Particularly, he proved the existence of negatively charged sub-atomic particles that are called as electrons which together with a positively charged nucleus form an atomic unit.

Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup." Thomson also made the following conclusions from his experiment:

The cathode ray is composed of negatively-charged particles.
The particles must be as part of an atom because the mass of each particle is only ~ 1/2000 compared to one hydrogen atom mass.

Within the atoms of all elements, these subatomic particles are found.

Initially it was controversial, but later Thomson's discoveries were gradually accepted by scientists. After then, his cathode ray particles were given a name called as electrons.

You might be interested in

What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me

alexgriva [62]

Concept: The magnification of spherical mirror can be defined by two ways.

(i) In terms of the height of the object and image.

The magnification of the spherical mirror is defined as the ratio of the height of the image' $h_{i}$ ' to the height of the object ' $h_{o}$ '. It is denoted by letter 'm'.

Mathematically, it can be written as

$m= \frac{h_{i}}{h_{o}} ------------(1)$

(ii) In terms of the object's and image's distances.

The magnification of the spherical mirror is defined as the negative ratio of the image distance' $d_{i}$ ' to the object distance ' $d_{o}$ '.

Mathematically, it can be written as

$m= - \frac{d_{i}}{d_{o}} ------------(2)$

Now, from equation (1) and (2) we have,

$m = \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} -----------(3)$

Given: Spherical Concave Mirror,

We will consider positive sign for object's and image's distance because both are in front of the mirror.

Object distance $(d_{o}) = + 8 mm$ .

Image distance $(d_{i}) = + 16 mm$

Object's height $(h_{o}) = + 4 mm$

Image's height $(h_{i}) =?$

Now, apply equation (3)

$\frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}}$

$Or, \frac{h_{i}}{4 mm} = - \frac{+16 mm}{+8 mm}$

Or, hi = - 8 mm

Here; negative sign means, the image will be inverted.

The image's height will be 8 mm.

4 0

3 years ago

Plz i need help for the 5 problems. plz show the work!!!

Artemon [7]

Answer:

1. 3 $m/s^{2}$

2. 1.5 $m/s^{2}$

3. 3 seconds

4. 0 $m/s^{2}$

5. 2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = $\frac {30-0}{10}=3 m/s^{2}$

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=10m/s, v=22m/s and t=8 seconds

a = $\frac {22-10}{8}=1.5 m/s^{2}$

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

$t=\frac {v-u}{a}$

Substituting u=0m/s since at rest, v=15m/s and a=5 $\frac {m}{s^{2}}$

= $\frac {15-0}{5}=3s$

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = $\frac {20-20}{10}=0 m/s^{2}$

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

$t=\frac {v-u}{a}$

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 $\frac {m}{s^{2}}$

= $\frac {0-9}{-4.1}=2.2s$

8 0

3 years ago

Find a first-degree polynomial function p1 whose value and slope agree with the value and slope of f at x = c. f(x) = cot(x), c

Neporo4naja [7]

The correct answer is y=-2x+(1/2)

y = f'(x)· x + c

Y = -2x + C

1 = -2x π/4 + C

=) C = I + π/2

y=-2x+(1/2) is the first-degree polynomial.

First-degree polynomials are the simplest polynomials. Here, we'll talk about a few qualities and connect the terms polynomial, function, and equation. Write a polynomial equation in standard form before attempting to solve it. Factor it, then set each variable factor to zero after it has reached zero. The original equations' answers are the solutions to the derived equations. Factoring cannot always be used to solve polynomial equations. For instance, the polynomial 2x+5 has an exponent of 1. The most typical kinds of polynomials used in algebra and precalculus are zero polynomial functions.

Learn more about polynomial functions here :-

brainly.com/question/22592200

#SPJ4

7 0

2 years ago

A runner has an original velocity of 6 m/s and slows to a final velocity of 0 m/s. If the runner covers a

myrzilka [38]

Answer:

4 s

Explanation:

Given:

Δx = 12 m

v₀ = 6 m/s

v = 0 m/s

Find: t

Δx = ½ (v + v₀) t

12 m = ½ (0 m/s + 6 m/s) t

t = 4 s

7 0

3 years ago

5 meddi muttasıl örnek

alexandr1967 [171]

Answer:

WHAT'S YOUR LANGUAGE

I CAN'T UNDERSTAND

5 0

3 years ago

Read 2 more answers