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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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If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the 1200.-kg pick-up truck behind him continues tr
Alborosie

Answer: 20.2 m/s

Explanation:

From the question above, we have the following data;

M1 = 800kg

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3 years ago
Read 2 more answers
Jane, looking for tarzan, is running at top speed 5.3 m/s and grabs a vine hanging vertically from a tall tree in the jungle. ho
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v₀ = initial speed as tarzan grabs the vine = 5.3 m/s

v = final speed as the tarzan reach the maximum height = 0 m/s

h = maximum height gained by the tarzan

m = mass of tarzan

using conservation of energy

initial kinetic energy = final kinetic energy + potential energy

(0.5) m v²₀ = (0.5) m v² + m g h

(0.5) v²₀ = (0.5) v² + g h

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3 0
3 years ago
A car accelerates at a rate of 13m/s^2[S]. If the car's initial velocity is 120km/h[N]. What will be its final velocity in m/s,
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Answer:

the final velocity of the car is 59.33 m/s [N]

Explanation:

Given;

acceleration of the car, a = 13 m/s²

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duration of the car motion, t = 2 s

The final velocity of the car in the same direction is calculated as follows;

v = u + at

where;

v is the final velocity of the car

v = 33.33 + 13 x 2

v = 59.33 m/s [N]

Therefore, the final velocity of the car is 59.33 m/s [N]

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