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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.

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Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above

poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

$U_{g,1} = U_{g,2} + W_{dis}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$W_{dis}$ - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

$U = m \cdot g \cdot y$

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$y$ - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

$W_{dis} = f \cdot \Delta s$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

$m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s$

$f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}$

If $m = 1000\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 40\,m$ , $y_{2} = 25\,m$ and $\Delta s = 400\,m$ , then:

$f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}$

$f = 367.763\,N$

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0

3 years ago

A 100.0 gram sample of Polonium-210 is contained for 552 days. How many half-lives occur during this period of time, if the half

Leto [7]

Answer:

4 half-lives will occur during this period of time.

Explanation:

Formula used :

$a_o=a\times e^{-\lambda t}$

$\lambda =\frac{0.693}{t_{1/2}}$

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives and time t

$a_o$ = Initial amount of the reactant.

$\lambda =$ decay constant

$t_{1\2}$ = half life of an isotope

n = number of half lives

We have :

$a_o=100.0 g$

a = ?

t = 552 days

$t_{1/2}=138 days$

$a=100.0 g\times e^{-\frac{0.693}{138}\times 552}$

$a=6.254 g$

$6.254 g=\frac{100.0 g}{2^n}$

$2^n=\frac{100.0 g}{6.254 g}$

n = 4

4 half-lives will occur during this period of time.

4 0

3 years ago

An unbanked circular highway curve on level ground makes aturn

tino4ka555 [31]

Answer:

Explanation:

Given

Velocity of traffic $v=60\ mi/hr\approx 26.82\ m/s$

Maximum value of Centripetal force is one-tenth of weight

such that $(F_c)_{max}=\frac{mg}{10}$

to make a safe turn centripetal force with radius of curvature r is given by

$F_c=\frac{mv^2}{r}$

$(F_c)_{max}=\frac{mg}{10}=F_c=\frac{mv^2}{r}$

$\frac{mg}{10}=\frac{mv^2}{r}$

$r=\frac{10v^2}{g}$

$r=\frac{10\times 26.82^2}{9.8}$

$r=733.99\approx 734\ m$

4 0

3 years ago

Explain in terms of the arrangementbof particles the kinetic theory of matter

baherus [9]

Answer:

Matter is made up of particles that are constantly moving. All particles have energy, but the energy varies depending on the temperature the sample of matter is in. This in turn determines whether the substance exists in the solid, liquid, or gaseous state.

Explanation:

5 0

3 years ago

What is the magnitude of the electric field strength between them, if the potential 7.05 cm from the zero volt plate (and 2.95 c

posledela

Answer:

E = 4156.02 Vm⁻¹

Explanation:

The magnitude of the uniform electric field between the plates can be given by the following formula:

$E = \frac{\Delta V}{d}\\$

where,

E = Electric field strength = ?

ΔV = Potetial Difference = 293 V

d = distance between plates = 7.05 cm = 0.0705 m

Therefore,

$E = \frac{293\ V}{0.0705\ m}\\\\$

6 0

3 years ago