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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.

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What type of speed looks at a particular point in time?

bearhunter [10]

The answer is D because instantaneous means at a particular point in time

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3 years ago

S When an uncharged conducting sphere of radius a is placed at the origin of an x y z coordinate system that lies in an initiall

telo118 [61]

The sphere has a constant potential. It is the electric field.

$E = V_{0} = 0$

In the sphere, then

$E_{x} = 0, E_{y}=0, E_{z}=0$

Outside the sphere, then

$V = V_{0} - E_{0}z + \frac{E_{0}a^{3}z}{(x^{2} +y^{2} + z^{2})^{3/2} }$

The elements of the electric field include

$E_{x} =\frac{3E_{0}a^{3}xy}{(x^{2} +y^{2} +z^{2})^{5/2}}\\E_{y} = \frac{3E_{0}a^{3}xz}{(x^{2} +y^{2}+z^{2})^{5/2}}$

Which becomes,

$=E_{0} (1-\frac{a^{3}}{x^{2} +y^{2}+z^{2})^{3/2}}+\frac{3a^{3}z^{2}}{(x^{2} +y^{2}+z^{2})^{5/2}})$

<h3>In a consistent electric field, is force constant?</h3>

Similar to an ordinary object in the uniform gravitational field near the Earth's surface, a charged item in a uniform electric field experiences a constant force and consequently experiences a uniform acceleration. The vector cross product of p and E determines the torque's direction.

If the charge is positive, the force either moves in the same direction as E or in the opposite direction (if charge is negative).

A torque is experienced by an electric dipole (p) in an even electric field (E). The vector cross product of p and E determines the torque's direction.

To learn more about uniform electric field, visit

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5 0

1 year ago

If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step

azamat

Answer:

Explanation:

$v_f^2 = v_i^2-2a(x_f-x_i)$

Subtract both sides by $v_i^2$ :

$- v_i^2+v_f^2 = -2a(x_f-x_i)$

Divide both sides by -2*a:

$\frac{v_i^2 - v_f^2}{2a} =x_f-x_i$

Add both sides by $x_i$ :

$x_i+\frac{v_i^2 - v_f^2}{2a} =x_f$

Subtract both sides by $\frac{v_i^2 - v_f^2}{2a}$ :

$x_i=x_f-\frac{v_i^2 - v_f^2}{2a}$

8 0

3 years ago

Hello help me pls! i need serious help

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Answer:c

Explanation:

7 0

2 years ago

Read 2 more answers

A generator produces 38 mwmw of power and sends it to town at an rms voltage of 78 kvkv. part a what is the rms current in the t

4vir4ik [10]

The rms current in the transmission lines is I = 487.18 A.

The root-imply-rectangular (rms) voltage of a sinusoidal supply of electromotive force is used to represent the source. it is the rectangular root of the time average of the voltage squared.

Alternating-present day circuits. the root-imply-square (rms) voltage of a sinusoidal source of electromotive force is used to symbolize the supply. it's far the square root of the time average of the voltage squared.

Electric power is by using present day or the waft of electric fee and voltage or the capacity of rate to deliver electricity. A given cost of power can be produced by using any combination of contemporary and voltage values

power = 38 M watt

rms voltage = 78 K v

power = IV

I = power/V

I = (38 * 1000000)/78*1000

I = 487.18 A.

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2 years ago