6. Give an example of an energy transformation from nuclear to electromagnetic.

1 answer:

7 0

An example of an energy transformation from nuclear to electromagnetic is: The sun transforms nuclear energy into ultraviolet, infrared, and gamma energy all forms of electromagnetic energy.

Explanation: The change of energy from one form to another is known as energy transformation. In every second energy transformations is constantly occurring everyplace.

Energy is transferable but it cannot be generated or destroyed. Electrical, mechanical, electromagnetic, nuclear, and chemical are some forms of energy.

The energy stored in the nucleus of an atom is known as Nuclear energy. Huge amounts of energy gets released at the time when nuclei combines or breaks. The gravitational collapse of hydrogen clouds produce heat and pressure that helps in activating Nuclear Fusion.

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)) What do these two changes have in common?

zloy xaker [14]

Answer:

Both are only physical changes

Explanation:

A physical change is a change that does not involve or alter the chemical composition of the substances involved. Physical changes form no new substance and can be easily separated into individual constituents. Example of physical changes are change in state, boiling, melting etc.

According to this question, two processes were given as follows:

1. mixing chocolate syrup into milk

2. rain forming in a cloud

These two processes are similar in the sense that they are both examples of physical changes.

7 0

2 years ago

25 PTS

yarga [219]

The answer would be C

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3 years ago

What is the strength of the electric field ep 0.90 mm from a proton?

dem82 [27]

The electric field strength is inversely related to the square of the distance.so the strength of the electric field is given by

$E=\frac{1}{4\pi \epsilon _{0} } \frac{q}{r^{2} } = \frac{k q}{r^{2} }$

Here, $\frac{1}{4\pi \epsilon _{0} } = k$ is constant depend upon medium and its value is $9.0 \times10^{9} \ N m^2/C^2$ and q is charge and r is the distance.

Given $r = 0.90 mm = 9.0 \times 10^{-4} m$ and we know the charge of proton, $q = 1.6\times 10^{-19} \ C$ .

Therefore,

$E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\ E= 1.77 \times 10^{-3} N/C$