B low frequency it is the lowest frequency
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
Answer:
Induced emf, 
Explanation:
Given that,
Length of the helicopter, l = 4 m
Angular speed of the helicopter, 
The vertical component of the Earth’s magnetic field is, 
We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :



So, the induced emf between the tip of a blade and the hub is
. Hence, this is the required solution.
Answer:
false. it's positive terminal of an electric cell
Energy that transfers through the medium