All of that fluff at the beginning is interesting, but completely irrelevant
to the question. The question is just asking for the mass of an object
that weighs 3.6N on Earth.
Weight = (mass) x (acceleration of gravity)
3.6N = (mass) x (9.8 m/s²)
Divide each side
by 9.8 m/s : Mass = 3.6N / 9.8 m/s² = <em>0.367 kilogram</em> (rounded)
Answer:
-30 °C
Explanation:
First, we have to calculate the molality (m) of the solution. If the solution is 50% C₂H₆O₂ by mass. It means that in 100 g of solution, the are 50 g of solute (C₂H₆O₂) and 50 g of solvent (water).
The molar mass of C₂H₆O₂ is 62.07 g/mol. The moles of solute are:
50 g × (1 mol / 62.07 g) = 0.81 mol
The mass of the solvent is 50 g = 0.050 kg.
The molality is:
m = 0.81 mol / 0.050 kg = 16 m
The freezing-point depression (ΔT) can be calculated using the following expression.
ΔT = Kf × m = (1.86 °C/m) × 16 m = 30 °C
where,
Kf: freezing-point constant
The normal freezing point for water is 0°C. The freezing point of the radiator fluid is:
0°C - 30°C = -30 °C
Looks like you simply substitute the length of the femur
Answer:
The angular acceleration α = 14.7 rad/s²
Explanation:
The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod
Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.
So Iα = Wr
Substituting the value of the variables, we have
mL²α/12 = mgL/2
Simplifying by dividing through by mL, we have
mL²α/12mL = mgL/2mL
Lα/12 = g/2
multiplying both sides by 12, we have
Lα/12 × 12 = g/2 × 12
αL = 6g
α = 6g/L
α = 6 × 9.8 m/s² ÷ 4.00 m
α = 58.8 m/s² ÷ 4.00 m
α = 14.7 rad/s²
So, the angular acceleration α = 14.7 rad/s²
Answer:
distance/ kinetic
Explanation:
According to the work energy theorem, the work done by all forces is equal to the change in kinetic energy of the body.
So, As the force is applied in the same direction of the distance traveled,so only the kinetic energy of the body changes as after application of force, the speed of the body changes.
