The weight is 45 N, because the three chains hold the sign, and each contributes 15 N.
Notice that the mass would be the weight/acceleration of gravity, m = 45/9.8 kg. But they ask the weight (force, so Newtons)
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
Speed = (frequency) x (wavelength)
Speed = (19 per second) x (7 mm)
Speed = (19 x 7) (per second · mm)
<em>Speed = 133 mm/sec</em>
or you might want to write <em>Speed = 0.133 m/s</em> .
