A child walks 5.0 meters north, then 4.0 meters east, and finally 2.0 meters south. what is the magnitude of the resultant displ
acement of the child after the entire walk
2 answers:
Answer:
Resultant displacement of the child is 5 m.
Explanation:
It is given that, a child walks 5.0 meters north, then 4.0 meters east, and finally 2.0 meters south.
We have to find the magnitude of displacement of the child after the entire walk. The attached figure shows entire journey of the child. Displacement of an object is equal to the shortest path covered by it. Here, AM shows the net displacement. It can be calculated using Pythagoras theorem as :
Since, AB = 5 m So, OA = 3 m and BC = OM = 4 m
⇒ AM = 5 m
Hence, the resultant displacement of the child after the entire walk is 5 m.
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D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values
Explanation:
The electric charge of an object is a property of the object that is related to the ability of the object to experience/exert an electric force: if the object is electrically charge, then it is attracted or repelled by other electrically charged object.
The electric charge of an object depends on the amount of charged particles it has on it. In particular, the fundamental particles that carry electric charge are:
- Protons: they carry electric charge of +e
- Electrons: they carry electric charge of -e
Where "e" is the fundamental charge (). Therefore, one proton carry a charge of +e and one electron carry a charge of -e.
An electron is a fundamental particle: this means that it cannot be divided into smaller particles. This also means that it is not possible to remove part of the charge of the electron: in fact, it is said that electric charge exists only as discrete values, being a multiple of . Therefore, the correct statement is
D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values
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Answer:
Object should be placed at a distance, u = 7.8 cm
Given:
focal length of convex lens, F = 16.5 cm
magnification, m = 1.90
Solution:
Magnification of lens, m = -
where
u = object distance
v = image distance
Now,
1.90 =
v = - 1.90u
To calculate the object distance, u by lens maker formula given by:
u = 7.8 cm
Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.