Probably because of the drag coefficient and the density of the liquid.
Answer:
single replacement
Explanation:
In the question it says one element replaces another element which means there is only one replacement.
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
We have that the cricket’s displacement is

From the Question we are told that
Start position 5 meter mark
It jumps to the 10 meter mark.
it jumps back to the 6 meter mark.
Generally the equation for cricket’s displacement. is mathematically given as


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Newton's second law allows calculating the response for the person's acceleration while leaving the trampoline is:
-4.8 m / s²
Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body
F = m a
Where the bold letters indicate vectors, F is the force, m the masses and the acceleration
The free body diagram is a diagram of the forces without the details of the body, in the attached we can see the free body diagram for this system
-W = m a
Whera
is the trampoline force
Body weight is
W = mg
We substitute
- mg = ma
a =
Let's calculate
a = 
a = -4.8 m / s²
The negative sign indicates that the acceleration is directed downward.
In conclusion using Newton's second law we can calculate the acceleration of the person while leaving the trampoline is
-4.8 m / s²
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