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3 years ago

A β− and a β+ particle meet and interact within a tissue. Discuss what would be the likely outcome of this interaction.

Physics

1 answer:

Arturiano [62]3 years ago

3 0

When a β− and a β+ particle meet and interact they are annihilated.

Every matter has an antimatter. An antimatter of a particular particle is another particle that has exactly the same properties as the initial particle but carries an opposite charge.

If we look at β− and β+ particles, we will notice that they are matter and antimatter. The interaction of matter and antimatter leads to annihilation. The energy produced by annihilation of matter and antimatter is enough to destroy the tissue.

So, when β− and a β+ particle meet and interact they are annihilated.

Learn more: brainly.com/question/23266051

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a 0.0818 kg salt shaker on a rotating table feels an inward frictional force of 0.108 N when it is moving 0.333m/s. what is the

serg [7]

Answer:

The required radius of its motion is $0.084m$ .

Explanation:

The formula for calculating the required radius of its motion is given by

$F = (mv^2)/r$

Where m= mass 

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F=frictional force 

r = radius of its motion 

Then the required radius of its motion is given by

$r =(mv^2)/F$

Given that

mas =0.0818 kg 

Frictional force= 0.108 N 

Moving with Velocity of = 0.333 m/s 

radius of its motion = $\frac{[0.0818 kg \times (0.333 m/s)^2]}{0.108 N}$ 

Hence the required radius of its motion is r = $8.4 cm=0.084m$ 

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3 years ago

Which one of the following types of microscope gives the highest amount of magnification?

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4 years ago

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castortr0y [4]

Answer:

635 years old

Explanation:

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Distance of star from Earth = 6.1 × 10⁸m

Speed of light = 3 × 10⁸ m/s

We have distance and speed, let us calculate the time of travel of the light from the star to the earth.

Distance = speed × time

6.1 × 10⁸ = 3 × 10⁸ × time

$time = \frac{6.1 \times 10^{18}}{3 \times 10^8}$

In order to do the division above, we will divide the whole numbers normally, then we will apply the law of indices to the power that says:

Xᵃ ÷ Xᵇ = X⁽ᵃ⁻ᵇ⁾

$\therefore time = \frac{6.1 \times 10^{18}}{3 \times 10^8}\\= \frac{2.03 \times 10^{(18-8)}}{1} \\= 2.03 \times 10^{10}}\ seconds$

Next, we are told that there are 3.2 × 10⁷ seconds in a year.

∴ The number of years travelled by the light from the star:

$3.2\ \times 10^7\ seconds = 1\ year\\1\ second = \frac{1}{3.2\ \times 10^7} \\\therefore 2.03 \times 10^{10}\ seconds = \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}$

please note that:

2.03 × 10¹⁰ = 20300000000

3.2 × 10⁷ = 32000000

$\therefore \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}\\= \frac{20300000000}{32000000} \\\\= \frac{20300}{32} \\= 634.347\ years\\$

The closest answer in the option is 635 years, and we are short of this by some points due probably to approximations in the calculation.

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