I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.
I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:
Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>
</span>
</span>
</span>
<span>
Distance at Perihelion
(</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>
Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.
</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>
1/2 (50%) of that is 43.9845 Earth days
The average of the aphelion and perihelion distances is
1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
1/2 ( 0.466697 + 0.307499) = 0.387 098 AU
This also happens to be 1/2 of the major axis of the elliptical orbit.
C,d,or e you can use the process of elimination to decide...
We can’t see the following
Answer:
Explanation:
90 rpm = 90 / 60 rps
= 1.5 rps
= 1.5 x 2π rad /s
angular velocity of flywheel
ω = 3π rad /s
Let I be the moment of inertia of flywheel
kinetic energy = (1/2) I ω²
(1/2) I ω² = 10⁷ J
I = 2 x 10⁷ / ω²
=2 x 10⁷ / (3π)²
= 2.2538 x 10⁵ kg m²
Let radius of wheel be R
I = 1/2 M R² , M is mass of flywheel
= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .
1/2 πR⁴ x t x d = 2.2538 x 10⁵
R⁴ = 2 x 2.2538 x 10⁵ / πt d
= 4.5076 x 10⁵ / 3.14 x .1 x 7800
= 184
R= 3.683 m .
diameter = 7.366 m .
b ) centripetal accn required
= ω² R
= 9π² x 3.683
= 326.816 m /s²
