Answer:
0.68 kg-m²
Explanation:
F = Force applied by the muscle = 2615 N
r = effective perpendicular lever arm = 2.85 cm = 0.0285 m
α = Angular acceleration of the forearm = 110.0 rad/s²
I = moment of inertia of the boxer's forearm = ?
Torque is given as
τ = I α eq-1
Torque is also given as
τ = r F eq-2
using eq-1 and eq-2
r F = I α
(0.0285)(2615) = (110.0) I
I = 0.68 kg-m²
Answer:
0.98 g/m
Explanation:
Note: Since Tension and frequency are constant,
Applying,
F₁²M₁ = F₂²M₂............... Equation 1
Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.
make M₂ the subject of the equation
M₂ = F₁²M₁/F₂²............... Equation 2
From the question,
Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz
Substitute these values into equation 2
M₂ = 196²(0.31)/110²
M₂ = 0.98 g/m
It’s distance for my answer but tell me if anyone tells u this answer thanks
ANSWER:
Ca(OH)2 shows a decrease of solubility with the increase of temperature.
Hope it helps u! :)
