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:
the object's mass is 50 kg
Explanation:
We use Newton's second law to solve for the mass:
F = m * a , then m = F / a
In our case, the acceleration is the gravitational acceleration on the planet, and the force is the weight of the object on the planet. So we get:
m = w / a = 650 N / 13 m/s^2 = 50 kg
Then, the object's mass is 50 kg.
D. All of the above. When a wire loop is moved or rotated in a magnetic field, there is a change in magnetic flux which produces emf in wire loop and hence electric current is produced.
