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Question 1

Take the equation e=mc^2 and rewrite the equation to solve for the speed of light "C" use rational exponents instead of roots.

Question 2
Using the properties of exponents apply the rational exponet to the numerator and the denominator of the fraction from your answer in part 1 then rationalize the denominator.

Question 3
Rewrite the euation without writing it as a fraction

Question 4
using whichever equation you prefer(part4,5, or 6.) find the speed of light in meters per second if 3 kilogram mass od matter converts to 2.7·10
x^(17) joules

1 Answer

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Question 1:

Starting with the equation e=mc^2, we can solve for the speed of light "C" by isolating it on one side of the equation.

First, we can divide both sides of the equation by "m":

e/m = c^2

Then, we can take the square root of both sides of the equation:

√(e/m) = c

So the equation solved for the speed of light "C" is:

c = √(e/m)

Note that we can also use rational exponents to write the equation, as follows:

c = (e/m)^(1/2)

Question 2:

Using the equation we found in Question 1 using rational exponents, we can apply the properties of exponents to simplify the expression:

c = (e/m)^(1/2)

c = [(e^(1/2))/(m^(1/2))]

To rationalize the denominator, we can multiply both the numerator and the denominator by the conjugate of the denominator:

c = [(e^(1/2))/(m^(1/2))] * [(m^(1/2))/(m^(1/2))]

c = (e^(1/2) * m^(1/2)) / m

c = (em)^(1/2) / m

Question 3:

To rewrite the equation without writing it as a fraction, we can start from the original equation e=mc^2 and solve for "c" by dividing both sides by "m" and then taking the square root of both sides:

e/m = c^2

√(e/m) = c

So the equation without writing it as a fraction is:

c = √(e/m)

Question 4:

We can use the equation c = √(e/m) to find the speed of light in meters per second if a 3 kilogram mass of matter converts to 2.7·10^17 J of energy:

c = √(e/m)

c = √((2.7·10^17 J) / (3 kg))

c = √(9·10^16 m^2/s^2)

c = 3·10^8 m/s

Therefore, the speed of light is approximately 3·10^8 meters per second.

User KrazzyNefarious
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