Answer:
The capacite is C=5.32 uF using the equations of voltage and energy in capacitance
Explanation:
The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J
Using:
Voltage in this case is the energy dissipated so
Using the equation to find capacitance
F
C= 5.32 uF because u is the symbol for micro that is equal to
It is a completely false statement that in <span>any energy transformation, there is always some energy that gets wasted as non-useful heat. The correct option among the two options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.</span>
To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,
Here,
= Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,
At time , Induced emf is,
Therefore the magnitude of the induced emf is 10.9V
Answer:
5x10^-3
Explanation:
Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.
Hooke's Law can be represented as
<h3> F = kx, </h3>
<em>where F is the force </em>
<em> k is the spring constant</em>
<em> x is the extension of the material </em>
<em />
Plug values in the equation
Step 1 find the original extension
0.045 = (400)x
x = 1.125x 10^-4 m d
Step 2 find the new extension
0.045+2 = 400(x)
2.045 = 400x
x = 5.1125x10^-3
Step 3 subtract the new extension with original
Total extension of the spring = 5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3