#18). (I think. It's the one that starts with "Compare..." Gravity ALWAYS attracts. The force between electric charges can attract or repel ... it depends whether the charges are the same kind or opposite kinds.
#19). With both gravity and electric charges, the force between them quickly becomes weaker when the distance between them increases.
#20). I don't think it changes. If the doorknob gets charged by something that TOUCHES it, so that charges can flow into it from the other object or out of it, then the total amount of electric charge on it might change. But the question says that the doorknob is charged by an "electric field", so nothing touched it, and charges couldn't flow into it or out of it. The only way it got charged was by the charges it already had in it getting moved around ... electrons in one part of the knob moving over to the other side. Then it would act as if it was charged ... if you touched it, you might get zapped.
#21)., #22)., #23). You're supposed to draw a graph to answer these. It's a very easy graph to draw, and you should do it. Label the x-axis 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Label the y-axis 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2 . Put the four points on the graph ... A, B, C, and D. For each point, the 'battery voltage' is the number on the x-axis and the 'Current' is the number on the y-axis. Then draw a line through the points. When you have the graph to look at, you can easily answer 21, 22, and 23.
24). I'm not sure, and I don't want to guess.
25). Did you ever move a coil of wire near a magnet in class ? This is the same situation, only the magnet is moving and the wire is still. The result will be the same.
26). Magnets have two poles that attract the opposite kind and repel. Now you copy the map and fill in the other side.
This is a lot of work for 5 points, so I left some parts for you to do. Another reason I did that is: You'll learn a lot more that way.
Its weight W is acting in downward direction. The upthrust force U acts in upward direction. When the balloon is moving upward, the air resistance is in downward and vice versa.
In this case, the hot-air balloon descends vertically at constant speed.
so,
and
so, ....................(1)
when it is ascending let the weight that it is releasing is R, so
..........(2)
solving equation (1) and (2)
2F is the weight of material that must be released from the balloon so that it ascends vertically at the same constant speed.
When the object is at the focal point the angular magnification is 2.94.
Angular magnification:
The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.
Here we have to find the angular magnification when the object is at the focal point.
Focal length = 6.00 cm
Formula to calculate angular magnification:
Angular magnification = 25/f
= 25/ 8.5
= 2.94
Therefore the angular magnification of this thin lens is 2.94