Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds
A) 1.55
The speed of light in a medium is given by:
where
is the speed of light in a vacuum
n is the refractive index of the material
In this problem, the speed of light in quartz is
So we can re-arrange the previous formula to find n, the index of refraction of quartz:
B) 550.3 nm
The relationship between the wavelength of the light in air and in quartz is
where
is the wavelenght in quartz
is the wavelength in air
n is the refractive index
For the light in this problem, we have
Therefore, we can re-arrange the equation to find , the wavelength in air:
Answer:
• They depend solely on the load that generates it
• Two or more electrical charges interact, which can be positive or negative
• The energy source is based on the electrical voltage
Answer:
0 m/s²
Explanation:
The velocity is constant, so there is no acceleration.
