Eh not really sure bout this one
<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as
Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as
where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then
Resolving for F,
PART B) We need here to apply the shear strength equation, then
In such a way that the material is more resistant to tensile strength than shear force.
Answer: Local governments can resolve a conflict by cooperating and spreading the costs of expensive projects. What kinds of problems cause state governments to help local governments? No economic growth or improvements are problems that cause governments to help local governments.
Explanation:
Show that the motion of a mass attached to the end of a spring is SHM
Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed
at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.
If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.
According to "Hook's Law
F = - Kx ---- (1)
Negative sign indicates that the elastic restoring force is opposite to the displacement.
Where K= Spring Constant
If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion
from a to b and then b to a.
According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by
F = ma ---- (2)
Comparing equation (1) & (2)
ma = -kx
Here k/m is constant term, therefore ,
a = - (Constant)x
or
a a -x
This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.
