A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time interval between ticks of the spacecraft clock to be 3.2 s. What is the time interval (in seconds) that an earth observer measures

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 rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .