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
You multiply the mass by the acceleration 82*7.5=615; that's what I would do
Answer:
114.32195122 but Round your answer to three significant figures.) is 114
Explanation:
Just took the test
Answer:
15.5 m/s.
Explanation:
Potential energy of the balloon has been converted to kinetic energy.
potential energy = kinetic energy.
mgh = ½mv².
10* 10* 12= ½ *10 *v²
1200 = 5v²
v²=1200÷5
v=√240
v= 15.49~15.5 m/s.
