5.91(approx) seconds just divide velocity by acceleration
Answer:mass of the object,how much force earth exerts on the object,and shape of the object
Explanation:
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: 37.5 nm
Explanation: speed of light c= 3.00·10^8 m/s.
I use same accuracy to speed of light as it's for frequency.
Frequency f= 8.01·10^15 1/s
Speed c = wavelength · frequency
Wavelength = c/f = 3.745·10^-8 m
