Question

The dimensionless parameter is used frequently in relativity. As y becomes larger and larger than 1, it means relativistic effects are becoming more and more important. What is y if v = 0.932c? A. 0.546 B. 0.362 C. 2.76 D. 3.83 E. 7.61

Answer 1

Answer:

$\gamma=2.76$

Explanation:

The relativistic parameter in theory of relativity is given by :

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

Where

v is the speed

c is the speed of light

Given that, v = 0.932 c

Put the value of v from above equation in equation (1) as :

$\gamma=\dfrac{1}{\sqrt{1-\dfrac{(0.932c)^2}{c^2}}}$

$\gamma=2.758$

or

$\gamma=2.76$

So, the value of the dimensionless parameter is 2.76. Hence, this is the required solution.