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3 years ago

14

This version of Einstein’s equation is often used directly to find what value?

2 answers:

Setler [38]3 years ago

8 0

This version of Einstein’s equation is often used directly to find what value? E = ∆mc2

Answer: This version of Einstein’s equation is often used directly to find the mass that is lost in a fusion reaction. Therefore the correct answer to this question is answer choice C).

I hope it helps, Regards.

Sindrei [870]3 years ago

5 0

<u>Answer</u>

A. the energy that is released in a nuclear reaction

<u>Explanation</u>

Robert Einstein devised the equation in 1905. He believed that some radioactive elements such as uranium constantly converted some of their mass to energy.

From the reactions of radioactive elements there came the production of nuclear energy. This energy is given by the Einstein’s equation, E = mc².

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The scanner records the time from when a ultrasound wave is emitted to when its reflection is received

telo118 [61]

Explanation:

The scanner records the time from when a ultrasound wave is emitted to when its reflection is received. A technician calculates the depth of the reflection using the equation as :

$\text{depth}=\dfrac{1}{2}\times \text{speed of ultrasound in patient}\times \text{time recorded by scanner}$

The distance covered by ultrasonic wave when it was emitted and gets reflected is 2d. Speed is given by :

$v=\dfrac{d}{t}$

d is distance and t is time

Here, d = 2d

So, the factor of (1/2) is used because the distance covered by the wave is 2 times when it was emitted and received.

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3 years ago

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Artyom0805 [142]

Answer:

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3 years ago

The more domains that are aligned in a magnet ___________.

I am Lyosha [343]

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3 years ago

A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv

sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

$K_{rel}=(\gamma-1)mc^2$

$\dfrac{K_{rel}}{mc^2}=\gamma-1$

Put the value into the formula

$\gamma=\dfrac{105.7}{0.511}+1$

$\gamma=208$

We need to calculate the electron’s velocity

Using formula of velocity

$\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}$

$\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}$

$\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1$

$v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2$

Put the value into the formula

$v^2=\dfrac{1-(208)^2}{-208^2}\times c^2$

$v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}$

$v=0.9999 c\ m/s$

Hence, The electron’s velocity is 0.9999 c m/s.

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3 years ago

(06.02 MC)<br> What can a scientist use to observe very small things?

Naddika [18.5K]

Answer:

microscopic light or Nano technology

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3 years ago