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andreev551 [17]
4 years ago
8

How was Einstein's theory that energy and mass are interchangeable eventually proven to be correct?

Physics
1 answer:
Pani-rosa [81]4 years ago
8 0

Answer:

Stanley Eddington confirmed the theory by his observations and inferences using the solar eclipse

Explanation:

Albert Einstein proposed a theory that energy and mass are interchangeable; termed the theory of special relativity. This can be observed when an object moves with the speed of light. The theory states that all laws in physics are the same for all non-accelerating observers.

Thus,

                      E = mc^{2

This theory could not be proven over many years, therefore it was not fully accepted.

In 1919, Stanley Eddington confirmed the theory by his observations and inferences using solar eclipse. The theory is now very accurate at any given speed when gravitational force tends to zero. This has lead to other observations to prove Einstein's theory.

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Answers:

a) t=0.311 s

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For this situation we will use the following equations:

y=y_{o}+V_{o}t+\frac{1}{2}at^{2} (1)  

V=V_{o} + at (2)  

Where:  

y is the <u>height of the model rocket at a given time</u>

y_{o}=0 is the i<u>nitial height </u>of the model rocket

V_{o}=0 is the<u> initial velocity</u> of the model rocket since it started from rest

V is the <u>velocity of the rocket at a given height and time</u>

t is the <u>time</u> it takes to the model rocket to reach a certain height

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<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

V=\frac{V_{1}+V_{2}}{2} (3)

For this rocket is:

V=\frac{27 m/s}{2}=13.5 m/s (4)

Time is determined by:

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Hence:

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Using equation (1), with initial height and velocity equal to zero:

y=\frac{1}{2}at^{2} (8)  

We will use y=4.2 m :

4.2 m=\frac{1}{2}a(0.311)^{2} (9)  

Finding a:

a=86.847 m/s^{2} (10)  

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again y=\frac{1}{2}at^{2} but for t=0.2 s:

y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2} (11)

y=1.736 m (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

V= at (13)  

V= (86.847 m/s^{2})(0.2 s) (14)  

Finally:

V=17.369 m/s (15)  

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