Can someone help me with these questions plz
1 answer:
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By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.
Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .
now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.
charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.
Now , use vector for solve it.
vector = vector Fe + vector Fn,
⇒ || =
⇒ Fe = Fs = KqQ/(1m)² = KqQ
⇒ = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}
⇒ = √2KqQ
Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.
To learn more about positive charges here
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Answer:
0.76
Explanation:
we are given:
radius (r) =5.7 m
speed (s) = 1 revolution in 5.5 seconds
acceleration due to gravity (g) = 9.8 m/s^{2}
coefficient of friction (Uk) = ?
we can get the minimum coefficient of friction from the equation below
centrifugal force = frictional force
m x r x ω^{2} = Uk x m x g
r x ω^{2} = Uk x g
Uk =
where ω (angular velocity) =
= = 1.14
Uk = = 0.76