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andre [41]
3 years ago
10

An air bubble of volume 20 cm³ is at the bottom of a lake 40 m deep, where the temperature is 4.0°C. The bubble rises to the sur

face, which is at a temperature of 20°C.Take the temperature of the bubble’s air to be the same as that of the surrounding water. Just as the bubble reaches the surface, what is its volume?
Physics
1 answer:
soldi70 [24.7K]3 years ago
6 0

Answer:

100 cm³

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.

n and R are constant, so:

P₁V₁/T₁ = P₂V₂/T₂

If we say point 1 is at 40m depth and point 2 is at the surface:

P₂ = 1.013×10⁵ Pa

T₂ = 20°C + 273.15 = 293.15 K

P₁ = ρgh + P₂

P₁ = (1000 kg/m³ × 9.8 m/s² × 40 m) + 1.013×10⁵ Pa

P₁ = 4.933×10⁵ Pa

T₁ = 4.0°C + 273.15 = 277.15 K

V₁ = 20 cm³

Plugging in:

(4.933×10⁵ Pa) (20 cm³) / (277.15 K) = (1.013×10⁵ Pa) V₂ / (293.15 K)

V₂ = 103 cm³

Rounding to 1 sig-fig, the bubble's volume at the surface is 100 cm³.

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Image result for if your going 30 m/s to go somewhere thats 1680 miles away
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1 year ago
A train travels 120 km in 2 hours and 30 minutes. What is its average speed?
ivann1987 [24]

Answer:

13.33 or 13 1/3m/s (meters per second)

Explanation:

In physics, we use the basic units of meters and seconds. So first convert (km) into meters (m) and also hours and minutes into seconds (s). We end up with 120000m and 9000s. Then divide the 120000m by the 9000s and you end up with 13.33 or 13 1/3 m/s.

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a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r
dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

brainly.com/question/13754413

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