Answer:
1/3
Explanation:
We can solve the problem by using the lens equation:
where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
Here we have a divering lens, so the focal length must be taken as negative (-f). Moreover, we know that the object is placed at a distance of twice the focal length, so
So we can find q from the equation:
Now we can find the magnification of the image, given by:
Answer:
5..No work is being done.
Explanation:
Hello!
Remember that the concept of work is defined as the force required to move a certain body distance, it is calculated as the product of force by distance.
therefore, the work required to raise the box is 150J.
However, the work required to keep the box lifted without moving is zero since although the box has a force due to the weight it does not move.
Answer:
Explanation:
Check attachment for solution
The correct answer D: all of the above
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.
