Answer:
x = 4,138 m
Explanation:
For this exercise, let's use the rotational equilibrium equation.
Let's fix our frame of reference on the left side of the pivot, the positive direction for anti-clockwise rotation
∑ τ = 0
n₁ 0 - W L / 2 + n₂ 4 - W_woman x = 0
x = (- W L / 2 + 4n2) / W_woman
Let's reduce the magnitudes to the SI System
M = 6 lbs (1 kg / 2.2 lb) = 2.72 kg
M_woman = 130 lbs = 59.09 kg
Let's write the transnational equilibrium equation
n₁ + n₂ - W - W_woman = 0
n₁ + n₂ = W + W_woman
n₁ + n₂ = (2.72 + 59.09) 9.8
At the point where the system begins to rotate, pivot 1 has no force on it, so its relation must be zero (n₁ = 0)
n₂ = 605,738 N
Let's calculate
x = (-2.72 9.8 6/2 + 4 605.738) / 59.09 9.8
x = 4,138 m
Answer is 3. Volume= mass divided by density
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
Answer:
1327 kg
Explanation:
So the net force exerted on the wagon would be the sum of forces from 2 horses subtracted by friction force

This force results in an acceleration of a = 1.3 m/s2. We can use Newton's 2nd law to calculate the mass of the wagon

