Please. Internet test due soon.

[The Motif]

Homework Statement

You observe a lead brick with a rest mass of 27 kg. What is its apparent mass if it is moving away from you at 0.73c? The speed of light is 3e8 m/s. Answer in units of kg.

What if you are moving toward it at -0.73c?

Homework Equations

The Attempt at a Solution

My Physics teacher hasn't taught us any equations yet to help solve these types of problems. He relies on us to use an "online calculator" that is currently not working on my computer.

 

[dynamicsolo]

You will want to search for the formula for "relativistic mass".

 

[ogb p]

As a scientist, it is important to have a strong understanding and knowledge of the equations and principles involved in solving problems like this. While relying on online calculators can be helpful, it is important to also have a solid understanding of the underlying concepts. In this case, the apparent mass of an object can be calculated using the equation:

m_apparent = m_rest / sqrt(1-(v^2/c^2))

Where m_rest is the rest mass of the object, v is the velocity of the object, and c is the speed of light.

Using this equation, we can calculate the apparent mass of the lead brick as it moves away from us at 0.73c. Plugging in the given values, we get:

m_apparent = 27 kg / sqrt(1-(0.73c)^2/(3e8m/s)^2) = 27 kg / sqrt(1-0.5329) = 27 kg / 0.8479 = 31.85 kg

Therefore, the apparent mass of the lead brick when it is moving away from us at 0.73c is approximately 31.85 kg.

Similarly, if we are moving towards the lead brick at -0.73c, the apparent mass can be calculated as:

m_apparent = 27 kg / sqrt(1-(-0.73c)^2/(3e8m/s)^2) = 27 kg / sqrt(1-0.5329) = 27 kg / 0.8479 = 31.85 kg

Which is the same as the apparent mass when the brick is moving away from us. This is because the apparent mass depends on the relative velocity between the observer and the object, not the direction of the velocity.

In summary, as a scientist, it is important to have a strong understanding of the underlying principles and equations involved in solving problems, rather than relying solely on online calculators. This will not only help in solving problems, but also in understanding and interpreting the results.