The answer is true! yes
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Answer:
1.35 m
Explanation:
Taking down to be positive, given:
Δx = Δy / tan 30.0º
v₀ₓ = 4.50 m/s
v₀ᵧ = 0 m/s
aₓ = 0 m/s²
aᵧ = 10 m/s²
Find: Δy
First, find the time it takes to land in terms of Δy.
Δy = v₀ t + ½ at²
Δy = (0 m/s) t + ½ (10 m/s) t²
Δy = 5t²
Next, find Δx in terms of t.
Δx = v₀ t + ½ at²
Δx = (4.50 m/s) t + ½ (0 m/s) t²
Δx = 4.50t
Substitute:
Δy = 5 (Δx / 4.50)²
20.25 Δy = 5 (Δx)²
4.05 Δy = (Δx)²
4.05 Δy = (Δy / tan 30.0º)²
4.05 Δy = 3 (Δy)²
1.35 = Δy
The basketball was thrown from an initial height of 1.35 m.
Graph: desmos.com/calculator/ujuzdo9xpr
What is it asking? It is sort of blurry
<span>Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116
So coefficient of static friction = 0.6116</span>
