Answer:
Q1 = 7.25*10^(-16) C
Explanation:
We are given;
electric field strength = (1 x 10^5 N/C
drag force (F) = 7.25 x 10^(-11) N
The question says it's moving with constant velocity. This means that he particle is in equilibrium and not accelerating.
Columbs law force of attraction or repulsion between two charges is given as;
F=(KQ1Q2)/r²
Now, electric field strength is given as the formula;(K*Q2)/r², thus plugging the relevant values gives us;
7.25 x 10^(-11) N= (1 x 10^(5) N/C)Q1 Q1 = 7.25 x 10^(-11) /(1 x 10^(5))
Q1 = 7.25*10^(-16) C
Answer:
An electric fan is considered to be a mixture of several simple machines. It includes the Wheel and Axle type, wedges, and the Inclined plane types. The blades of an electric fan are the inclined planes and the wedges.
Explanation :
Simple machines makes our work easier. Lever is one of the simple machine which consists of rigid rod that is pivoted at a fixed support called as Fulcrum.
There are three classes of lever.
Class 1 : In this type of class, fulcrum is placed in between effort and load. Hence the movement of load is in reverse direction of the movement of effort. (fig 1)
Class 2 : In this type of, the load is between the effort and the fulcrum. Hence, the movement of load is in same direction as that of the effort. (fig 2)
Class 3 : In this type of lever the effort between the load and the fulcrum. Hence, both the effort and load are in same direction. (fig 3)
Hence, when the position of fulcrum is modified the effort force changes.
Answer:
-45,597.07
Explanation:
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