What is the equivalent capacitance of the three capacitors in the figure?
1 answer:
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Answer:
Guessing you just need help with the definition but if it's the question I can still help you.
We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:
Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:
Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:
Now, we divide by the distance of the muscle:
Now, we substitute the values:
Now, we solve the operations:
Therefore, the force exerted by the muscles is 23.53 Newtons.
Part B. To determine the force on the pivot we will add the forces we add the vertical forces:
Since there is no vertical movement the sum of vertical forces is zero:
Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:
Now, we plug in the values:
Solving the operations:
Therefore, the force is 73.53 Newtons.
Answer:
The electric charge, q (in coulomb units) = 5004 C
Given:
The charge stored as printed on NiMH battery, q = 1390 mAh
Solution:
To express the amount of electric charge printed on the battery in milli-ampere-hour (mAh) in coulomb, we will do simple conversion of milli amperes in ampere and hours in seconds:
1 mA =
1 hour =
Also, we know that the rate of flow of charge is electric current, I:
I =
⇒ q = [tex]I\times t[tex] (1)
where
q = electric charge
I = current
t = time taken for flow of current
Using eqn(1), we get:
q = [tex]1390\times 10^{-3}\times 60\times 60[tex]
q = 5004 A-s = 5004 C