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.
magnetic field due to a finite straight conductor is given by
here since it forms an equilateral triangle so we will have
also the perpendicular distance of the point from the wire is
now from the above equation magnetic field due to one wire is given by
now since in equilateral triangle there are three such wires so net magnetic field will be
1.7 x 10^2 N
or 166 N
First you find the vertical component of the weight, which is 9.8*40, (g*m), which is 392 N. You then find the angle between that and the slope, which is 90-25, which is 65. You then multiply the vertical weight by cos(65), to find the component of that that is parallel to the slope. You get 165.666 N
Answer:
The positively charged ball moves between both charged plates till the plates and the ball all become neutral.
Check Explanation for more.
Explanation:
Let the ball be in square brackets, and the plates in normal brackets.
(+) [+] (-)
From the law that like charges repel and unlike charges attract.
The positive ball would go first to the negatively charged plate. After which, the ball would hold more negative charges overall than before.
Because the ball is now more negatively charged, it then travels towards the positive plate. In the same manner, the ball would transfer negative electrons to the positive plate.
So, when leaving the positive plate, the ball would be more positive and be drawn towards the negative plate once more. In doing so, it would make the negative plate more positive.
Then, the ball again holds more negative electrons and is drawn towards the positive plate once more.
This back and forth process continues until the once-positive and once-negative plates become neutral, that is, they are discharged.
The ball hanging on the insulated thread becomes neutral too at this point.
Hope this Helps!!!
The period T of a pendulum is given by:
where L is the length of the pendulum while
is the gravitational acceleration.
In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is
. Using this data, we can solve the previous formula to find L:
