Answer:
θ = 29.38°
Explanation:
The centripetal force is given by the formula;
F_c = F_n(sin θ) = mv²/r
Now, the vertical component of the normal force is; F_n(cos θ)
Now, this vertical component is also expressed as; F_n(cos θ) = mg
Thus, the slope is;
F_n(sin θ)/F_n(cos θ) = (mv²/r)/mg
tan θ = v²/rg
v² = rg(tan θ)
The initial speed will be gotten from the relation;
(v_o)² = μ_s(gr)
Plugging rg(tan θ) for (v_o)², we have;
μ_s(gr) = rg(tan θ)
rg will cancel out to give;
μ_s = (tan θ)
Thus, θ = tan^(-1) μ_s
μ_s is coefficient of static friction given as 0.563
θ = tan^(-1) 0.563
θ = 29.38°
The answer is 5kg m/s with the second momentum being 10 kg m/s. i took the test and it was right. hope this helps yalls
Answer: vf= 51 m/s and d= 112 m
Explanation: solution attached
