Answer:
Capacitive Reactance is 4 times of resistance
Solution:
As per the question:
R =
where
R = resistance
f = fixed frequency
Now,
For a parallel plate capacitor, capacitance, C:
where
x = separation between the parallel plates
Thus
C ∝
Now, if the distance reduces to one-third:
Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.
Also,
Also,
Z ∝ I
Therefore,
Solving the above eqn:
The distance between slit and the screen is 1.214m.
To find the answer, we have to know about the width of the central maximum.
<h3>How to find the distance between slit and the screen?</h3>
where, d is the distance between slit and the screen, and a is the slit width.
Thus, we can conclude that, the distance between slit and the screen is 1.214m.
Learn more about the width of the central maximum here:
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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:
Explanation:
The formula for this, the easy one, is
where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:
and simplifying a bit:
and
N = 48.0(.0625) so
N = 3 mg left after 12.3 years