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3 years ago

How did J.J. Thomson's experiments demonstrate that Dalton's model of the atom was incorrect?

2 answers:

6 0

<span>Dalton's famous experiments using cathode ray tubes demonstrate the presence of electrons, of which Dalton was previously unaware. In his theories, Dalton posited that atoms were solid, indivisible particles. However, Thomson's experiments showed that there were in fact subatomic particles, ushering in a new era of study of the atom.</span>

morpeh [17]3 years ago

4 0

The correct option is C.
J.J Thompson demonstrated that Dalton's model of the atomic theory was wrong by showing that atoms are made up of sub particles, this is contrary to the report that Dalton gave about atoms. In his experiment, Dalton reported that atoms are indivisible, that is, they can not be broken down into smaller particles. J.J Thompson on the other was able to show through his cathode ray experiment that atoms are made up of smaller particles called electrons.

You might be interested in

Which two statements about kinetic energy are true

gtnhenbr [62]

Answer:

A, C

Explanation:

It can be found in objects like a swing. It creates kinetic energy when it is coming back or going away, while moving.

8 0

2 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

3 0

2 years ago

PLEASE PLEASE HELP!

vichka [17]

Well, the relationship between the net force and mass and acceleration of an object are directly related, as per the equation - Fnet = ma.

Thus the solution is A. As the net force of an object decreases, the object's acceleration also decreases, mass is kept constant.

5 0

4 years ago

Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied

Anettt [7]

Answer:

loloeuhsh

Explanation:

shwvwgnwajejjeisus

5 0

3 years ago

200 mL of an ideal gas is placed in a piston and is held at a pressure of 500 torr. If the temperature is held constant and the

galina1969 [7]

Answer:

V₂=153.84 m L

Explanation:

Given that

Initial volume ,V₁ = 200 m L

Initial pressure ,P₁= 500 torr

Final pressure P₂ = 650 torr

Lets take the final volume =V₂

Given that process occurs at constant temperature and we know that for constant temperature process

P₁ V₁ = P₂ V₂

$V_2=\dfrac{P_1V_1}{P_2}$

Now by putting the values

$V_2=\dfrac{500\times 200}{650}\ mL$

V₂=153.84 m L

Therefore the new volume of the gas will be 153.8 4mL.

7 0

3 years ago