To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as
Here,
k = Spring constant
m = Mass
Our values are given as,
Rearranging to find the spring constant we have that,
Therefore the spring constant is 1.38N/m
This is false. Current is the speed of the charge, 1 amp of current is 1 coulomb per second. So you can imagine the current of a circuit as the current of a river. In a parallel circuit, the river breaks into two separate streams. Some of the water goes down one river, some goes down the other. However, the total amount of water/coulombs never changes. This means that some of the total current will go down one river, and one the other. However, with less coulombs now the current will decrease.
Long story short, since there are two paths, the charge will split and depending on the resistance of each parallel stream a different amount of charge will go down each branch.
Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.
Ek = mv^2 / 2 — multiply both sides by 2
2Ek = mv^2 — divide both sides by m
2Ek / m = V^2 — switch sides
V^2 = 2Ek / m — plug in values
V^2 = 2*30J / 34kg
V^2 = 60J/34kg
V^2 = 1.76 m/s — sqrt of both sides
V = sqrt(1.76)
V = 1.32m/s (roughly)
Answer:
The force due to friction is generally independent of the contact area between the two surfaces. This means that even if you have two heavy objects of the same mass, where one is half as long and twice as high as the other one, they still experience the same frictional force when you drag them over the ground.
Explanation:
Independent
E=Eq,
where F is the electrostatic force (or Coulomb force) exerted on a positive test charge q.
