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2 years ago

Which of the following types of particles can be categorized into different kind of elements?

Physics

2 answers:

KatRina [158]2 years ago

4 0

The answer would be atoms

antiseptic1488 [7]2 years ago

3 0

Answer: The correct answer is Atom.

Explanation:

Atom consists of proton, electron, neutron and nucleus. The subatomic particles of the atom are electron, proton and neutron.

The electrons revolve around the nucleus while neutrons and protons are located in nucleus. The whole mass of the atom is supposed to be concentrated at nucleus.

The electron is negatively charged particle. The proton is positively charged particle. The neutron has no charge. It is neutral.

Therefore, the correct option is B.

You might be interested in

(Select all that apply)

loris [4]

Answer: a. air pollution

c. hazardous wastes

d. potential reactor accident

e. water pollution

A nuclear energy is produced in a thermal power plant. A nuclear energy is produced in a nuclear reactor. In nuclear reactor nuclear fission reactions takes place in which an atoms absorbs energy from radiations and undergo fission and produces energy in the form of high intensity radiations along with heat. Although the fission reactions takes place in a nuclear reactor in a controlled way so that the radiation may not leak out from the reactor. The accidentally leak out radiations or explosion or bursting of the reactor due to uncontrolled thermal energy production can result in air pollution as the leak out air will cause bursting effects which will contaminate the air.

The nuclear waste are radioactive and are non-biodegradable these wastes are disposed off deep in geospheres and in water. They have potential to contaminate both land and water. Radioactive wastes can cause mutations in the genome of the organisms exposed to these wastes which generate deadly diseases and disorders. Therefore, these wastes are hazardous.

5 0

3 years ago

Read 2 more answers

An electron that has a velocity with x component 2.4 x 106 m/s and y component 3.6 x 106 m/s moves through a uniform magnetic fi

likoan [24]

Answer:

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force, a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as: $\overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\$

In respect to the magnetic field; $\overrightarrow{B}=(0.027 i - 0.15 j) [T]$

The magnetic force can be written as;

$\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]$

Bear in mind $q =-1.6021x10^{-19} [C]$

thus,

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.

6 0

3 years ago

Which of the following is not true about tectonic plates

AlladinOne [14]

Answer:

Scientists have identified about a dozen major and several minor tectonic plates

6 0

2 years ago

Tres personas, A, B, C, jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección ho

Yuri [45]

Answer:

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

Explanation:

Si la caja debe hallarse en equilibrio físico, entonces se debe satisfacer la siguiente ecuación:

$F_{A} + F_{B} + F_{C} = 0$ (1)

Si sabemos que $F_{A} = -3$ y $F_{B} = 5$ , entonces el valor de la fuerza que debe ejercer la persona C debe ser:

$F_{C} = -F_{A}-F_{B}$

$F_{C} = -(-3)-5$

$F_{C} = -2$

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

3 0

3 years ago

Suppose a current flows through a copper wire. Which two things occur?

Bingel [31]

Answer:

Explanation:

as the copper wire is very dangerous so now if these two thing happens then it would easily help the current flows through it so it might be a little bit easy for the current to flow through it

4 0

2 years ago