Volume = mass / density
Volume = 20 / 7.87
Volume = 2.54 (2 s.f)
Answer:
V = 4.48m/s a = 1.57m/s²
Explanation:
ω₀ = 16rad/s
α₀ = 5.6rad/s²
r = 280mm = 0.28m
a = ?
v = ?
Angular velocity (ω₀) = velocity of acceleration / length of path
ω₀ = v / r
V = ω₀ * r
V = 16 * 0.28
V = 4.48m/s
Acceleration = ?
Angular acceleration α₀ = angular velocity (ω) / time take (t)
α₀ = ω / t .... equation i
But acceleration (a) = velocity (v) / time (t)
a = v / t
t = v / a
Put t = v / a into equation i
α₀ = ω / (v / a)
α₀ = ω * a / v
α * v = ω * a
a = (α * v) / ω
a = (5.6 * 4.48) / 16
a = 1.568m/s²
a = 1.57m/s²
You haven't told us anything about the detectors being used. We don't know how the sensitivity of the detector is related to the total number of photons absorbed, and we don't even know whether you and your friend are both using the same type of detector.
All we can do, in desperation, is ASSUME that the minimum time required to just detect a star is inversely proportional to the total number of its photons that strike the detector. That is, assume . . .
(double the number of photons) ===> (detect the source in half the time) .
-- The intensity of light delivered to the prime focus of a telescope is directly proportional to the AREA of its objective lens or mirror, which in turn is proportional to the square of its radius or diameter.
So your telescope gathers (0.18/0.05)² = 12.96 times as much light as your friends telescope does.
-- So we'd expect your instrument to detect the same star in
(119.5 min) / (12.96) = <em>9.22 minutes .</em>
We're simply comparing the performance of two different telescopes as they observe the same object, so the star's magnitude doesn't matter.
Do you remember the formula for potential energy ?
PE = (mass) (gravity) (height)
If the mass and the height are always the same, then the least PE comes from the least gravity. Surely you can find THAT in the table.
