Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .
1 answer:
Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.
Explanation:
- Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
- Let
be the speed of the river's current given as 1.00 m/s.
- Note that this speed is the magnitude of the velocity which is a vector quantity.
- The direction of the swimmer is upstream.
Hence the resultant velocity is given as, = S — S 0
= 1.25 — 1
= 0.25 m/s.
Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.
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Answer:
When viewed from above, the current in the coil should point towards the top-right corner of the picture.
Explanation:
The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.
Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.
With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.
Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current . Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field
. The direction of the right thumb should now point in the direction of the force on the wire
.)
Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.
It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.
Answer:
The magnification of an astronomical telescope is -30.83.
Explanation:
The expression for the magnification of an astronomical telescope is as follows;
Here, M is the magnification of an astronomical telescope, is the focal length of the eyepiece lens and
is the focal length of the objective lens.
It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.
Put and
in the above expression.
M=-30.83
Therefore, the magnification of an astronomical telescope is -30.83.