Which of the following is a risk associated with taking performance-enchancing drugs.
1 answer:
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Answer:
= ( ρ_fluid g A) y
Explanation:
This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force
for the first part, let's write Newton's equilibrium equation
B₀ - W = 0
B₀ = W
ρ_fluid g V_fluid = W
the volume of the fluid is the area of the cube times the height it is submerged
V_fluid = A y
For the second part, the body introduces a quantity and below this equilibrium point, the equation is
B - W = m a
ρ_fluid g A (y₀ + y) - W = m a
ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a
ρ_fluid g A y + (B₀-W) = ma
the part in parentheses is zero since it is the force when it is in equilibrium
ρ_fluid g A y = m a
this equation the net force is
= ( ρ_fluid g A) y
we can see that this force varies linearly the distance and measured from the equilibrium position
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
Total potential due to this charges = 4500 V + 6750 V = 11250 V