Probably to the right because the force on the left is greater and will therefore overpower the 2N force and push to the right
Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
Answer:
The beat frequency is 30 Hz
Explanation:
Given;
velocity of the two sound waves, v = 343 m/s
wavelength of the first wave, λ₁ = 5.72 m
wavelength of the second wave, λ₂ = 11.44 m
The frequency of the first wave is calculated as follows;
F₁ = v/λ₁
F₁ = 343 / 5.72
F₁ = 59.97 HZ
The frequency of the second wave is calculated as follows;
F₂ = v/λ₂
F₂ = 343 / 11.44
F₂ = 29.98 Hz
The beat frequency is calculated as;
Fb = F₁ - F₂
Fb = 59.97 HZ - 29.98 Hz
Fb = 30 Hz
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