L = length of the incline = 75 m
θ = angle of incline = 22 deg
h = height of skier at the top of incline = L Sinθ = (75) Sin22 = 28.1 m
μ = Coefficient of friction = 0.090
N = normal force by the surface of incline
mg Cosθ = Component of weight of skier normal to the surface of incline opposite to normal force N
normal force "N" balances the component of weight opposite to it hence we get
N = mg Cosθ
frictional force acting on the skier is given as
f = μN
f = μmg Cosθ
v = speed of skier at the bottom of incline
Using conservation of energy
potential energy at the top of incline = kinetic energy at the bottom + work done by frictional force
mgh = f L + (0.5) m v²
mgh = μmg Cosθ L + (0.5) m v²
gh = μg Cosθ L + (0.5) v²
(9.8 x 28.1) = (0.09 x 9.8 x 75) Cos22 + (0.5) v²
v = 20.7 m/s
Answer:
Explanation:
When a liquid is heated, the particles are given more energy. They start to move faster and further apart. At a certain temperature, the particles break free of one another and the liquid turns to gas. This is the boiling point.
Answer:
α = -π/3 rad/s²
θ = 1.5π rad ≈ 4.71 rad
θ = 0.75 rev
Explanation:
30 rev/min (2π rad/rev) / (60 s/min) = π rad/s
α = (ωf - ωi) / t = (0 - π) / 3 = -π/3 rad/s²
θ = ½αt² = ½(π/3)3² = 1.5π rad ≈ 4.71 rad
θ = 1.5π rad / 2π rad/rev = 0.75 rev
Answer:
0.0129 m
Explanation:
ΔL = FL / (EA)
where ΔL is the deflection,
F is the force,
L is the initial length,
E is Young's modulus,
and A is the cross sectional area.
F = mg = 100 kg × 9.8 m/s² = 9800 N
A = 4.0 mm² × (1 m / 1000 mm)² = 4×10⁻⁶ m²
ΔL = (9800 N) (1.0 m) / ((1.9×10¹¹ Pa) (4×10⁻⁶ m²))
ΔL = 0.0129 m