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
B. truthfully it's the only one that makes sense
First, we calculate the mass of Phosphorous present:
283.88 x 0.4364
= 123.88 amu
Atomic mass of P is 31 amu
moles of P = mass / Ar
= 123.88 / 31
= 4.0 moles
We know that one mole of substance has 6.02 x 10²³ particles
Atoms of P = 4 x 6.02 x 10²³
= 2.41 x 10²⁴ atoms
True clicking the office button and then clicking new would display the new document.
Answer:
Explanation:
Assume that the distance travelled initially is d.
In order to stop the block you need some external force which is friction.
If we use the law of energy conservation:
a)
Looking at the formula you can see that the mass doesn't affect the distance travelled, as lng as the initial velocity is constant (Which indicates that the force must be higher to push the block to the same speed) therefore the distance is the same.
b) If the velocity is doubled, then the distance travelled is multiplied by 4, because the distance deppends on the square of the velocity.
