Answer:
9155 years old
Explanation:
We use the following expression for the decay of a substance:

So we first estimate the value of k knowing that the half-life of the C14 is 5730 years:

so, now we can estimate the age of the artifact by solving for"t" in the equation:

which we can round to 9155 years old.
-Synodic period is the period of celestial bodies observed on the moving planet(mostly earth)
Sideral period is the period comparing to the fixed stars without motion of the earth involved.
(I will explain the second question with an example, so it's easier to understand)
-For Sideral month for example of the moon it cactually complete one revolution in around 27.3 days.
However, since the earth moves, for us it took some more time to see the moon the same as before (fullmoon to fullmoon) again. That make synodic month of the moon to be around 29.5 days.
The magnitude of the induced emf is given by:
ℰ = |Δφ/Δt|
ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time
The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:
φ = BA
B = magnetic field strength, A = loop area
The area of the loop A is given by:
A = πr²
r = loop radius
Make a substitution:
φ = B2πr²
Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:
Δφ = ΔB2πr²
ΔB = change in magnetic field strength
Make another substitution:
ℰ = |ΔB2πr²/Δt|
Given values:
ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s
Plug in and solve for ℰ:
ℰ = |(-0.20)(2π)(0.50)²/2.5|
ℰ = 0.13V
Answer:
B. When the ball is released, the thrower's arm transfers its energy to the ball.