Answer: Load divided by it effort
Explanation:
Mechanical advantage of any machine is its load divided by its effort
<span> </span>For any prism-shaped geometry, the volume
(V) is assumed by the product of cross-sectional area (A) and height (h).
<span> V = Ah </span>
<span>
Distinguishing with respect to time gives the
relationship between the rates.
dV/dt = A*dh/dt</span>
<span> in the meantime the area is not altering </span>
<span>
dV/dt = π*(1 ft)^2*(-0.5 ft/min) </span>
<span>
dV/dt = -π/2 ft^3/min ≈ -1.571 ft^3/min
Water is draining from the tank at the rate of π/2
ft^3/min.</span>
Explanation:
a. KE at bottom = PE at top
½ mv² = mgh
v = √(2gh)
v = √(2 × 9.8 m/s² × 20.0 m)
v = 19.8 m/s
b. Work by friction = PE at top
mgμ d = mgh
d = h / μ
d = 20.0 m / 0.210
d = 95.2 m
Answer:
va = 4.79 m/s
vb = 1.29 m/s
Explanation:
Momentum is conserved:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(3.00) (0) + (6.50) (3.50) = (3.00) v₁ + (6.50) v₂
22.75 = 3v₁ + 6.5v₂
For an elastic collision, kinetic energy is conserved.
½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²
m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
(3.00) (0)² + (6.50) (3.50)² = (3.00) v₁² + (6.50) v₂²
79.625 = 3v₁² + 6.5v₂²
Two equations, two variables. Solve with substitution:
22.75 = 3v₁ + 6.5v₂
22.75 − 3v₁ = 6.5v₂
v₂ = (22.75 − 3v₁) / 6.5
79.625 = 3v₁² + 6.5v₂²
79.625 = 3v₁² + 6.5 ((22.75 − 3v₁) / 6.5)²
79.625 = 3v₁² + (22.75 − 3v₁)² / 6.5
517.5625 = 19.5v₁² + (22.75 − 3v₁)²
517.5625 = 19.5v₁² + 517.5625 − 136.5v₁ + 9v₁²
0 = 28.5v₁² − 136.5v₁
0 = v₁ (28.5v₁ − 136.5)
v₁ = 0 or 4.79
We know v₁ isn't 0, so v₁ = 4.79 m/s.
Solving for v₂:
v₂ = (22.75 − 3v₁) / 6.5
v₂ = 1.29 m/s
