When a trebuchet counterweight is hoisted by soldiers, what form of energy is
2 answers:
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This question is incomplete, the complete question is;
According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g
where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.
Do you think this equation is valid in any system of units
Answer:
YES, the equation is a general equation that is valid in any system of units
Explanation:
Given the data in the question;
h = (0.04 to 0.09)(D/d)⁴ ×
so
[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)
[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)
∴ [ L ] = (0.04 to 0.09) [L]
So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.
Therefore, YES, the equation is a general equation that is valid in any system of units
Answer:
t = 2 v₀ / g
Explanation:
For this projectile launch exercise we use the displacement equations
x = vox t
y = y₀ + t - ½ g t²
As it is launched horizontally the vertical velocity is zero and the point of origin of the coordinate system is here, so y₀ is zero.
x = v₀ t
y = ½ g t²
They ask us for the time for which
x = y
vo t = ½ g t²
t = 2 v₀ / g