1) We use:
2as = v² - u², with v = 0,
to find the acceleration of the arrow.
2 x 0.038 x a = -(25)²
a = -8.22 x 10³ m/s²
F = ma
F = 6.5 x 10⁻² x -8.22 x 10³
F = -534.2 N; the negative direction indicates that the force is in the opposite direction of the motion.
B) The arrows force is the same but in the opposite directioin.
534.2 N
C) a = -8.22 x 10³ m/s²
s = -(65)²/(2 x -8.22 x 10³)
s = 25.7 cm
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.
Answer:
1 Ohm is defined as the resistance of a conductor with a potential difference of 1 volt applied to the ends through which 1-ampere current flows. Ohms is the SI unit of electrical resistance
