Answer:
Its length is measured to be 0.5 m
Explanation:
From theory of relativity (mass variation), we know that:
m = mo/√(1-v²/c²)
Where, m = relative mass
and, mo = rest mass
The momentum of stick while moving, will be:
P = mv
but, it is given in the form of rest mass as:
P = 2(mo)v
thus, by comparison;
2(mo)v = mv
using value of m from theory of relativity;
2(mo)v = (mo)v/√(1-v²/c²)
√(1-v²/c²) = 1/2 ______ eqn(1)
Now, for relativistic length (L), we have the formula from same theory of relativity;
L = (Lo)√(1-v²/c²)
The rest length (Lo) of meter stick is 1 m, and the remaining term on right side √(1-v²/c²), known as Lorentz Factor, can be given by eqn (1), as equal to 1/2.
Thus,
L = (1 m)(1/2)
<u>L = 0.5 m</u>
Answer:
λ = 5940 Angstroms
Explanation:
This is an exercise of the relativistic Doppler effect
f’= f √((1- v / c) / (1 + v / c))
Where the speed in between the strr and the observer is positive if they move away
Let's use the relationship
c = λ f
f = c /λ
We replace
c /λ’ = c /λ √ ((1- v / c) / (1 + v / c))
λ = λ’ √ ((1- v / c) / (1 + v / c))
Let's calculate
v = 0.01 c
v = 0.01 3 10⁸
v= 3 10⁶ m / s
λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]
λ = 6000 √ [0.99 / 1.01]
λ = 5940 Angstroms
Answer:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
Explanation:
Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:
(1)
Where:
, 
- Initial and final momemtums of the small object, measured in kilogram-meters per second.
, 
- Initial and final momentums of the big object, measured in kilogram-meters per second.
If we know that 
, 
and 
, then the final momentum of the big object is:
The magnitude of the large object's momentum change is:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
Answer:
No
Explanation:
Because the people put in charge of defining SI units said so
