Answer:
30.96 m
Explanation:
If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:
d = v * Δt
v = 0.8*c = 0.8 * 3e8 = 2.4e8
Δt = ζ = 129 ns = 1.29e-7 s
d = 2.4e8 * 1.29e-7 = 30.96 m
This is the distance as measured by observer A.
Answer:
-2.5m/s²
Explanation:
The acceleration of a body is giving by the rate of change of the body's velocity. It is given by
a = Δv / t ----------------(i)
Where;
a = acceleration (measured in m/s²)
Δv = change in velocity = final velocity - initial velocity (measure in m/s)
t = time taken for the change (measured in seconds(s))
From the question;
i. initial velocity = 5m/s
final velocity = 0 [since the body (ball) comes to rest]
Δv = 0 - 5 = -5m/s
ii. time taken = t = 2s
<em>Substitute these values into equation (i) as follows;</em>
a = (-5m/s) / (2s)
a = -2.5m/s²
Therefore, the acceleration of the ball is -2.5m/s²
NB: The negative sign shows that the ball was actually decelerating.
Answer:
Hello your question has some missing parts attached below is the complete question
answer : 4 μm
Explanation:
since the scale bar works the same way as a scale on a map , each bar will therefore represent 1 μm and the mature parent cell's is about 4 times the labeled value hence the Mature parent cell diameter will approximately be : 4 μm
