g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb
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Answer:
v₁₀ = 1.90 m / s
Explanation:
In this exercise we are given the maximum height data, with energy we can know how fast the body came out
Final mechanical energy, maximum height
= U = m g h
Initial mechanical energy, in the lower part of the track
Em₀ = K = ½ m v²
Em=
½ m v² = m g h
v = √ 2gh
Now we can use the moment to find the speed with which objects collide
The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and final velocity v
Initial before the crash
p₀ = M v₁₀ + 0
Final after the crash
= M v1f + m v
p₀ =
M v₁₀ = M + m v
As the shock is elastic the kinetic energy is conserved
K₀ =
½ M v₁₀² = ½ M ² + ½ m v²
Let's write the system of equations
M v₁₀ = M + m v
M v1₁₀² = M ² + m v²
We cleared v1f in the first we replaced in the second
= (M v₁₀ - mv) / M
M v₁₀² = M (M v₁₀ - mv)² / M² + m v²
M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²
v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0
2 m v₁₀ - v (m + 1) m/ M = 0
v₁₀ = v (m +1) / (2M)
Let's substitute the value of v
v1₁₀= √ (2gh) (m +1) / (2M)
Let's calculate
v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2 5.41)
V₁₀ = 7.668 (2.68) / 10.82
v₁₀ = 1.90 m / s
Answer:
a. 0.000002 m
b. 0.00000182 m
Explanation:
36 cm = 0.36 m
15 cm = 0.15 m
a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:
Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same
Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:
As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:
b) If the temperatures changes, we can still reuse the ideal gas equation above: