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3 years ago

Which scientist thought that the atom was the smallest particle in the universe?

1 answer:

4 0

The scientist thought that the atom was the smallest particle in the universe is John Dalton. The answer is letter B. He established the atomic theory which consists of five; elements are made of extremely small particles called atoms, atoms of different element have different sizes, mass and physic – chemical properties, atoms cannot be divided further, destroyed or created, atoms can combine to form compounds and in chemical reaction, atoms can be combined, separated or rearranged. J.J. Thompson discovered the electrons in the atom. Ernest Rutherford discovered the half – life of an atom and Neils Bohr explain the quantum mechanics.

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Process of Science Task: Earth-Centered vs. Sun-Centered Models.

Gnoma [55]

Answer:

There are a number of models as explained below.

Explanation:

Depending on the position of the earth in the orbit, stars will be in different positions relative to the distant stars.

This model can be explained by the parallax effect. An object is viewed differently by the left and the right eye.

Reference can be made to the geocentric and heliocentric arguments. The first view argues that heavens are composed of 55 concentric spheres.

On the other hand, the heliocentric system, proposed by Copernicus, suggested that the Sun, not the earth, was the center of the solar system.

6 0

3 years ago

show answer Incorrect Answer 33% Part (b) Find the radius of curvature, in meters, of the path of a proton accelerated through t

timofeeve [1]

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

Answer: (a) V = - 109.44 x 10² V

(b) $r_{p}=$ 9.95 x 10⁻¹ m

Explanation: (a) Potential difference is defined as the energy a charged particle has between two points in a circuit. It is calculated as

$\Delta V=\frac{pe}{q}$

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = $\frac{1}{2}m.v^{2}$

$V.q=\frac{1}{2} m.v^{2}$

$V=\frac{m.v^{2}}{2q}$

Calculating:

$V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}$

V = $-109.44$ x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is $-109.44$ x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

$F_{mag} = F_{c}$

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

$r=\frac{mv}{qB}$

The radius of the curvature, for a proton, will be:

$r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

$r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 54.33 x 10⁻⁵m

ratio = $\frac{9.95.10^{-1}}{54.33.10^{-5}}$

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

5 0

4 years ago

How are atomic emission spectra like fingerprints for the elements

storchak [24]

Atomic emission spectra are like fingerprints for the elements, because it can show the number of orbits in that elements as well as the energy levels of that element. As each emission of atomic spectra is unique, it is the fingerprint of element.

Explanation:

Each element has unique arrangement of electrons in different energy levels or orbits. So depending upon the difference in energy of the orbital, the emission spectra will be varying for each element. As the binding energy and excitation energy is not common for any two elements, so the spectra obtained when those excited electrons will release energy to ground state will also be unique.

As in atomic emission spectra, the incident light will be absorbed by the electrons of those elements making the electron to excite, then the excited electron will return to ground state on emission of radiation of energy. Thus, this energy of emission is equal to the difference between the energy of initial and final orbital. So the spectra will act like fingerprints for elements.

8 0

3 years ago

A 60-watt light bulb has a voltage of 120 bolts applied across it and a current of 0.5 amperes flows through the bulb. What is t

Andre45 [30]

Answer:

240 ohms

Explanation:

From Ohms law we deduce that V=IR and making R the subject of the formula then R=V/I where R is resistance, I is current and V is coltage across. Substituting 120 V for V and 0.5 A for A then

R=120/0.5=240 Ohms

Alternatively, resistance is equal to voltage squared divided by watts hence $\frac {120^{2}}{60}=240$

7 0

4 years ago

Help me plssssss will give brainliest if correct

Lina20 [59]

Answer:

electricity is the answer

3 0

3 years ago

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