71.2k views
0 votes
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

3 votes

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
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories