If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.
If you apply 1 newton to the machine's input, the output force is
(2.5 x 1 newton) = 2.5 newtons.
If you apply 120 newtons to the machine's input, the output force is
(2.5 x 120 newtons) = 300 newtons.
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀
Answer:
A. 0.199 J
B. 0.0663 C
C = 0.0221 F
D. 12.68 ohms
Explanation:
From the question:
time duration, t = 0.28 seconds
Average power, P = 0.71 W
Average voltage, V = 3 V
A) Energy is given as:
E = P * t
=> E = 0.71 * 0.28 = 0.199 J
B) Electrical energy is also given as:
E = qV
where q = charge
=> q = E / V
∴ q = 0.199 / 3 = 0.0663 C
C) Capacitance is given as charge over voltage:
C = q / V
=> C = 0.0663 / 3 = 0.0221 F
D) Electrical power, P, can also be given as:
P = 
where R = resistance
=> R = 
R = 
Trial and error
scientific laws and theories are proven by experimental data and large bodies of evidence.