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NISA [10]
1 year ago
9

how does a particle differ from its anti-particle? it has opposite mass. it has opposite charge. its wave-function is the opposi

te sign. all of the above none of the above
Physics
1 answer:
FromTheMoon [43]1 year ago
7 0

An antiparticle is a subatomic particle that, by definition, has the same mass as its normal particle counterpart but the opposite magnetic moment and electric charge. An electron's antiparticle, for instance, is the positron.

The mass of an antiparticle is equal to that of the particle version, but it has the opposite charge. Antiparticles have opposite charges, baryon numbers, lepton numbers, and strangeness. A subatomic particle known as an antiparticle has the same mass as a particle of regular matter but the opposite electric charge and magnetic moment. Thus, the positron (a positively charged electron) is the opposite of the negatively charged electron.

To know more about electric charge,

brainly.com/question/11461182

#SPJ4

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It loses it's fuel supply from the water and begins to weaken
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Please help with this question! Thanks in advance...
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5 0
3 years ago
Is there any difference between reflection and echo???
Fofino [41]

Answer:

reflection is the act of reflecting or the state of being reflected while echo is

reflection of sound waves from a surface back to the initial listener

6 0
3 years ago
What type of decay has the release of electromagnetic energy from the nucleus and has the most penatrating power through other s
Verdich [7]

Answer:

Gamma decay

Explanation:

There are 3 types of radioactive decay:

- Alpha decay: in this decay, a nucleus emits an alpha particle (consisting of 2 protons and 2 neutrons, so a nucleus of helium). The alpha particle has a large charge (2e) and a large mass (4u), so it is strongly ionizing, and therefore loses energy faster while moving through matter, therefore its penetrating power is low (it can be easily stopped by a thin sheet of paper or by the skin)

- Beta decay: this decay occurs when a neutron in a nucleus turns into a proton, emitting a beta particle (a fast-moving electron) alongside with an antineutrino. The beta particle has a lower charge (e) and a smaller mass than the alpha particle, so it has a moderate penetrating power, being able to penetrate more than the alpha particle (the beta particle can be stopped by a thin sheet of aluminium)

- Gamma decay: this decay occurs when an excited nucleus decays emitting a gamma ray photon (which is electromagnetic energy with very high energy and frequency). The gamma ray photon has no charge and no mass, therefore it has the most penetrating power, being able to travel a much large distance before being absorbed by matter (several metres of concrete are required to stop gamma radiation.

So, the description in the question refers to gamma decay.

7 0
3 years ago
1) A car accelerates uniformly from rest and reaches a speed of 19.8 m/s in 7 s. The diameter of a tire is 38.5 cm. Find the num
andre [41]

Answer:

1) 57 revolutions.

2) 4.52 m

3) 1531.2 m

Explanation:

Question 1:

Given:

Initial velocity, u = 0

Final velocity, v = 19.8\ m/s

Time, t = 7\ s

Acceleration, a = \frac{\textrm{Final velocity-Initial velocity}}{Time}=\frac{19.8}{7}=2.83\ m/s^2

Now, displacement of the tire is given as:

S=ut+\frac{a}{2}t^2\\S=0+\frac{2.83}{2}7^2=69.335\ m

Displacement of tire in 1 revolution is equal to its circumference.

Therefore, displacement in 1 revolution = \pi\times(Diameter)=\pi \times 38.5\times 10^{-2}=1.2095\ m

Now, number of revolutions is given as:

N=\frac{Total\ displacement}{Displacement\ per\ revolution}\\N=\frac{69.335}{1.2095}=57

Therefore, the number of revolutions are 57.

Question 2:

Given:

Radius of the wheel is, R=4.8\ m

Angle of rotation is, \theta=54°

Converting degree to radians, we get:

\theta=54\times \frac{\pi}{180}=0.3\pi

Now, path length is given as:

L=R\theta=(4.8)(0.3\pi)=1531.2\ m

Therefore, the path length of a point on the wheel is 4.52 m

Question 3:

Radius of the wheel is, R=4.8\ m

Angle of rotation is, \theta=319 radians

Now, path length is given as:

L=R\theta=(4.8)(319)=4.52\ m

Therefore, the path length of a point on the wheel is 1531.2 m.

7 0
3 years ago
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