Answer: A (smooth)
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Answer:
34 m/s
Explanation:
Potential energy at top = kinetic energy at bottom + work done by friction
PE = KE + W
mgh = ½ mv² + Fd
mg (d sin θ) = ½ mv² + Fd
Solving for v:
½ mv² = mg (d sin θ) − Fd
mv² = 2mg (d sin θ) − 2Fd
v² = 2g (d sin θ) − 2Fd/m
v = √(2g (d sin θ) − 2Fd/m)
Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:
v = √(2 (9.8 m/s²) (150 m sin 28°) − 2 (50 N) (150 m) / (65 kg))
v = 33.9 m/s
Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.
Answer:
1.635×10^-3m
Explanation:
Young modulus is the ratio of the tensile stress of a material to its tensile strain.
Young modulus = Tensile stress/tensile strain
Tensile stress = Force/Area
Given force = 130N
Area = Πr² = Π×(1.55×10^-3)²
Area = 4.87×10^-6m²
Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²
Tensile strain = extension/original length
Tensile strain = e/3.9
Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²
2×10¹¹N/m² = 8.39×10^7/{e/3.9)}
2×10¹¹ = (8.39×10^7×3.9)/e
2×10¹¹e = 3.27×10^8
e = 3.27×10^8/2×10¹¹
e = 1.635×10^-3m
The stretch of the steel wire will be
1.635×10^-3m
