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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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WARRIOR [948]

Answer:

As we need to use a nested loop in our function,hence push $ra

pop $ra

jal nested_function_label

nop is the correct option.

8 0
4 years ago
A bomb of mass 6kg initially at rest explodes into two fragments of masses 4kg and2kg respectively. If the greater mass moves wi
Katen [24]

Answer:

v = 10 [m/s]

Explanation:

The largest mass is that of 4 [kg], in this way the momentum can be calculated by means of the product of the mass by velocity.

P=m*v\\

where:

P = momentum [kg*m/s]

m = mass = 4 [kg]

v = velocity = 5 [m/s]

Now the momentum:

P=4*5\\P=20[kg*m/s]

This same momentum is equal for the other mass, in this way we can find the velocity.

P=m*v\\20=2*v\\v=10[m/s]

7 0
3 years ago
A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm
andrew11 [14]

Answer:

2.8 cm

Explanation:

y_1 = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm

Fringe width is also given by

\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}

For second order

\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1

Distance between two second order minima is given by

y_2=2\beta_2

\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm

The distance between the two second order minima is 2.8 cm

8 0
3 years ago
How long must a tow truck apply a force of 600 N to increase the speed of a 1,500 kg car at rest to 2 m/s?
slega [8]

The time the truck must apply the given force to increase its speed to given value is 5 s.

The given parameters;

  • <em>applied force, F = 600 N</em>
  • <em>mass of the truck, m = 1,500 kg</em>
  • <em>speed of the truck, v = 2 m/s</em>

The force applied to the truck is determined by Newton's second law of motion; <em>which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.</em>

F = ma

F = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{1500 \times 2}{600} \\\\t = 5 \ s

Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.

Learn more here:brainly.com/question/1988795

7 0
3 years ago
At what angle are the electronic and the magnetic wave related in an electromagnetic signal?
VikaD [51]

Answer:

90degrees I'm pretty sure

7 0
3 years ago
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