Respuesta :
Answer:
Speed of the object, v = 0.99 c
Explanation:
It is given that, an object has a relativistic momentum that is 8.30 times greater than its classical momentum.
The relativistic momentum is given by, [tex]p=\dfrac{m_ov}{\sqrt{1-\dfrac{v^2}{c^2}}}[/tex]
Classical momentum, p' = mv
According to the given condition,
p = 8.3 p'
[tex]\dfrac{m_ov}{\sqrt{1-\dfrac{v^2}{c^2}}}=8.3\times m_ov[/tex]
[tex]\sqrt{1-\dfrac{v^2}{c^2}}=\dfrac{1}{8.3}[/tex]
[tex]1-\dfrac{v^2}{c^2}=0.01451[/tex]
[tex]\dfrac{v^2}{c^2}=0.985[/tex]
[tex]v=0.99c[/tex], c is the speed of light
So, the speed of the object is 0.99 c. Hence, this is the required solution.
Answer:
v = .992 c.
Explanation:
In relativistic mechanics , momentum
[tex]mv=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2} } }[/tex]
c is velocity of light , m₀ is rest mass and v is velocity of the body..
Given
[tex]\frac{mv}{m_0v} = 8.3[/tex]
[tex]\sqrt{1 -\frac{v^2}{c^2} } = \frac{1}{8.3}[/tex]
Solving for v
v = .992 c.
