The force exerted by one plate for two parallel plates arrangement is F = QE/2.
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Force one plate exerts on the other</h3>
The force exerted by one plate for two parallel plates arrangement is given as follows;
F = QE/2
where;
- Q is the charge on one plates
- E is the electric field due to the charge
- F is force exerted by one plate
Thus, the force exerted by one plate for two parallel plates arrangement is F = QE/2.
Learn more about force between parallel plates here: brainly.com/question/13590045
<span><span><span>FIRST EQUATION OF MOTION
Vf = Vi + at</span> </span><span>Consider a body initial moving with velocity "Vi". After a certain interval of time "t", its velocity becomes "Vf". Now</span>Change in velocity = Vf - Vi <span>
OR
DV =Vf – Vi</span><span>Due to change in velocity, an acceleration "A" is produced in the body. Acceleration is given by</span>a = DV/t Putting the value of "DV"<span><span>a = (Vf – Vi)/t
at = Vf – Vi
at + Vi =Vf
</span>OR
</span><span>Vf = Vi + at
</span>
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Because many objects in space don't radiate any optical (visible) radiation at all.
And other objects, like stars, radiate a lot of invisible radiation in addition to the
visible light from them. So the ability to detect and measure invisible radiation
makes it possible to learn a lot more about objects in space than we could if
we could only use their visible light.
To solve this problem it is necessary to apply the concepts related to the magnetic flow of a coil and take into account the angles for each case.
It is also necessary to delve into part C, the concept of electromotive force (emf) which is defined as the variation of the magnetic flux as a function of time.
By definition the magnetic flux is determined as:
Where
N = Number of loops We will calculate the value for each of the spins
B = Magnetic Field
A = Cross-sectional Area
Angle between the perpendicular cross-sectional area and the magnetic field.
PART A) The magnetic flux through the coil after it is rotated is as follows:
PART B) For the second case the angle formed is perpendicular therefore:
PART C) The average induced emf of the coil is as follows:
Answer:
a)= 29.4J
b)F = 588 N
c)= 60 Kg
Explanation:
Force constant of the spring (k) = 5880 N/m
Change in length of the spring (x) = 25 - 15 = 10 cm 0.1m
This work done on the spring as it is stretched (or compressed) can be recovered. This is stored work that can be used to do work on something else by this spring. That means the stretched (or compressed) spring has energy -- potential energy. This is spring potential energy or elastic potential energy.
a) Work done in pulling the body W = 1/2kx²
= 1/2 (5880)(0.1)2
= 29.4J
b)From Hook's Law,
F = ke
Where F = applied force, k = spring constant, e = extension.
Given: k = 5880 N/m, e = 25-15 = 10 cm = 0.1 m.
Substitute into the formula above
F = 5880(0.1)
F = 588 N
c)By using the formula, F = -kx
Hence mg = kx
Thus m x 9.8 =5880 x0.1
Hence mass of the body
m= 5880 x0.1/9.8
= 60 Kg