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lara31 [8.8K]
3 years ago
14

A spherical soap bubble with a surface-tension of 0.005 lbf/ft is expanded from a diameter of 0.5 in to 3.0 in. How much work, i

n Btu, is required to expand this bubble
Physics
1 answer:
Sladkaya [172]3 years ago
8 0

Answer:W = 1.23×10^-6BTU

Explanation: Work = Surface tension × (A1 - A2)

W= Surface tension × 3.142 ×(D1^2 - D2^2)

Where A1= Initial surface area

A2= final surface area

Given:

D1=0.5 inches , D2= 3 inches

D1= 0.5 × (1ft/12inches)

D1= 0.0417 ft

D2= 3 ×(1ft/12inches)

D2= 0.25ft

Surface tension = 0.005lb ft^-1

W = [(0.25)^2 - (0.0417)^2]

W = 954 ×10^6lbf ft × ( 1BTU/778lbf ft)

W = 1.23×10^-6BTU

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A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin
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To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

\Delta L = L_0 \alpha \Delta T

Where,

L_0 = Initial Length

\alpha = Thermal expansion coefficient

\Delta T = Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C

103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C

The coefficient of thermal expansion given is

\alpha = 1.1*10^{-5}/\°C

The initial length would be,

L_0 = 565m

Replacing we have to,

\Delta L = L_0 \alpha \Delta T

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = 0.355m

This means that the building will be 35.5cm taller

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3 years ago
The form of energy in fossil fuels is what kind of energy?
Nonamiya [84]
Solar energy or sunlight...
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How do i find the number of leptons in an atom??
Serjik [45]

Answer:

In particle physics, a lepton is an elementary particle of half-integer spin (spin 1⁄2) that does not undergo strong interactions.[1] Two main classes of leptons exist: charged leptons (also known as the electron-like leptons or muons), and neutral leptons (better known as neutrinos). Charged leptons can combine with other particles to form various composite particles such as atoms and positronium, while neutrinos rarely interact with anything, and are consequently rarely observed. The best known of all leptons is the electron.

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A satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther a
irinina [24]

The speed of the second satellite is less than the speed of the first satellite.

<h3>What is speed?</h3>

The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.

Given is a satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther away from the same planet.

When the distance from the center of the orbit increases, the time to complete the orbit will be greater.

Thus, the speed of the second satellite is less than the speed of the first satellite.

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brainly.com/question/7359669

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3 0
2 years ago
. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim
inn [45]

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ \frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}

On multiplying both sides by  "100", we get

⇒ \frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}

⇒ \frac{\Delta T}{T}\times 100=0.005787 (%)

∵  T=2\pi\sqrt{\frac{l}{g} }

On substituting the values, we get

⇒ \frac{\Delta T}{T}% = \frac{1}{2}\times \frac{\Delta l}{l}%

On applying cross multiplication, we get

⇒ \frac{\Delta l}{l}% = 2\times \frac{\Delta T}{T}%

⇒        = 2\times 0.05787

⇒        = 0.011574

⇒        = 0.012%

6 0
3 years ago
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