University:
Florida Gateway College
Course:
MAC 2312 | Calculus II
Academic year:

2024
Views:

449

Pages:

2
Author:

Ruby Bates

Question 1 Fraction of an iceberg above the surface of the water: The specific gravity of ice is 0.917, and that of seawater is 1.025. Solution: Using Archimedes' principle: Weight of ice = Weight of displaced seawater Weight of ice=Weight of displaced seawater Volume submerged = 𝜌 ice 𝜌 seawater = 0.917 1.025 ≈ 0.895 Volume submerged= ρ seawater ρ ice = 1.025 0.917 ≈0.895 The fraction of the iceberg above the surface: Fraction above = 1 − 𝜌 ice 𝜌 seawater = 1 − 0.895 = 0.105 Fraction above=1− ρ seawater ρ ice =1−0.895=0.105 Answer: About 10.5% of the iceberg is above the surface. Question 2 Speeds of two pucks after a perfectly elastic collision: Given: Mass of puck 1: 𝑚 1 = 0.450 kg m 1 =0.450kg, 𝑣 1 = 3.00 m/s v 1 =3.00m/s, Mass of puck 2: 𝑚 2 = 0.900 kg m 2 =0.900kg, 𝑣 2 = 0 m/s v 2 =0m/s. The final velocities for elastic collisions are: Velocity of puck 1: 𝑣 1 ′ = 𝑚 1 − 𝑚 2 𝑚 1 + 𝑚 2 𝑣 1 = 0.450 − 0.900 0.450 + 0.900 ( 3.00 ) = − 1.00 m/s v 1 ′ = m 1 +m 2 m 1 −m 2 v 1 = 0.450+0.900 0.450−0.900 (3.00)=−1.00m/s Velocity of puck 2: 𝑣 2 ′ = 2 𝑚 1 𝑚 1 + 𝑚 2 𝑣 1 = 2 ( 0.450 ) 0.450 + 0.900 ( 3.00 ) = 2.00 m/s v 2 ′ = m 1 +m 2 2m 1 v 1 = 0.450+0.900 2(0.450) (3.00)=2.00m/s Answer: Speed of puck 1: 1.00 m/s (opposite direction), Speed of puck 2: 2.00 m/s (same direction as puck 1's original velocity). Question 3 Power required to operate the ski tow: Given: Slope 𝜃 = 1 5 ∘ θ=15 ∘ , Length 𝐿 = 300 m L=300m, Speed 𝑣 = 12.0 km/h = 3.33 m/s v=12.0km/h=3.33m/s, Riders: 𝑁 = 50 N=50, Average rider mass 𝑚 = 70.0 kg m=70.0kg. Step 1: Work against gravity Gravitational force: 𝐹 = 𝑁 ⋅ 𝑚 ⋅ 𝑔 ⋅ sin ⁡ 𝜃 = 50 ⋅ 70.0 ⋅ 9.8 ⋅ sin ⁡ ( 1 5 ∘ ) F=N⋅m⋅g⋅sinθ=50⋅70.0⋅9.8⋅sin(15 ∘ ) 𝐹 = 50 ⋅ 70.0 ⋅ 9.8 ⋅ 0.2588 ≈ 8891 N F=50⋅70.0⋅9.8⋅0.2588≈8891N Step 2: Power calculation Power: 𝑃 = 𝐹 ⋅ 𝑣 = 8891 ⋅ 3.33 ≈ 29 , 600 W P=F⋅v=8891⋅3.33≈29,600W Answer: The power required is approximately 29.6 kW. Question 4 Largest radius for a car to remain in contact with a loop: Given: Speed 𝑣 = 4.0 m/s v=4.0m/s, Gravitational force provides centripetal force at the top of the loop. At the top of the loop: 𝑚 𝑔 = 𝑚 𝑣 2 𝑅 mg= R mv 2 𝑅 = 𝑣 2 𝑔 = ( 4.0 ) 2 9.8 ≈ 1.63 m R= g v 2 = 9.8 (4.0) 2 ≈1.63m Answer: The largest radius is approximately 1.63 m.