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hammer [34]
3 years ago
5

A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of

14 cm^3. The bubble rises to the surface, where the temperature is 23 degrees C. What is the volume of the bubble (in cubic centimeters) just before it reaches the surface? (The density of water is 1000 kg/m^3 and air pressure is 1E+05 N/m^2.)
Physics
1 answer:
posledela3 years ago
6 0

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

By substituting the values, we get

\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

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The Mariana trench is located in the Pacific Ocean at a depth of about 11 000 m below the surface of the water. The density of s
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Answer:

A) 27209506.5 N

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The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.

Explanation:

Force du to depth of water is

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P = density of salt water = 1025 kg/m3

g = acceleration due to gravity 9.81 m/s2

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3 years ago
It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s 2 . Find the magnitude of acce
shepuryov [24]

This Question is not complete

Complete Question:

a. A hawk flies in a horizontal arc of radius 11.3 m at a constant speed of 5.7 m/s. Find its centripetal acceleration.

Answer in units of m/s2

b. It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s2. Find the magnitude of acceleration under

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Answer in units of m/s2

Answer:

a. 2.875m/s²

b. 3.172m/s²

Explanation:

a. The formula for centripetal acceleration = (speed²) ÷ radius

Centripetal acceleration = (5.7m/s)²÷ 11.3m

Centripetal acceleration = 2.875m/s²

b. Magnitude of acceleration can be calculated by finding the sum of the vectors for the both the centripetal acceleration and the increase in the speed rate.

Centripetal acceleration ( acceleration x) = 2.875m/s²

Increase in the speed rate ( acceleration n) = 1.34m/s²

Magnitude of acceleration = √a²ₓ + a²ₙ

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= √ 10.06m/s²

= 3.172m/s²

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