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
a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):
It is given:
v = 8.5 m/s
m = 91 kg
So:
b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h
In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J
m = 91 kg
g = 9.81 m/s² - gravitational acceleration
h = ? - height
Now, let's replace those:
3,287.4= 91 * 9.81 * h
⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m
Answer: d₂ = 170 mGya
Explanation:
the relationship between absonbed 'd' and exposure 'E' is given as;
D(Gv) = F . x (AS/xB)
F is a conversion coefficient depending on medium
so we can simply write
d₁/d₂ = x₁/x₂
Given that;
our x₁ = 60 mAs, x₂ = 120 mAs, d₁ = 85 mGya, d₂ = ?
from the given formula,
d₂ = (x₂d₁ / x₁)
now we substitute
d₂ = (120 × 85) / 60
d₂ = 170 mGya
∴ if 120 mAa is used, the new exposure will be 170 mGya
