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stich3 [128]
3 years ago
11

A spherical balloon is inflating with helium at a rate of 72 ft2/min . How fast is the​ balloon's radius increasing at the insta

nt the radius is 3 ​ft? How fast is the surface area​ increasing?
Physics
1 answer:
Yanka [14]3 years ago
7 0

The volume of the balloon is given by:

V = 4πr³/3

V = volume, r = radius

Differentiate both sides with respect to time t:

dV/dt = 4πr²(dr/dt)

Isolate dr/dt:

dr/dt = (dV/dt)/(4πr²)

Given values:

dV/dt = 72ft³/min

r = 3ft

Plug in and solve for dr/dt:

dr/dt = 72/(4π(3)²)

dr/dt = 0.64ft/min

The radius is increasing at a rate of 0.64ft/min

The surface area of the balloon is given by:

A = 4πr²

A = surface area, r = radius

Differentiate both sides with respect to time t:

dA/dt = 8πr(dr/dt)

Given values:

r = 3ft

dr/dt = 0.64ft/min

Plug in and solve for dA/dt:

dA/dt = 8π(3)(0.64)

dA/dt = 48.25ft²/min

The surface area is changing at a rate of 48.25ft²/min

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The force of gravity is the force with which massively large objects such as the earth attracts another object towards itself. All objects of the earth exert a gravity that is directed towards the center of the earth. Therefore, the force of gravity of the earth is equal to the mass of the object times acceleration due to gravity and also equal to the weight of the object.

 

F = ma

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Answer:

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Explanation:

We are given that

Radius,r=0.3 m

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According to question

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200m

Explanation:

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¿Cuáles de las siguientes cualidades permiten identificar un cuerpo como planeta? I) Debe ser aproximadamente esférico. II) Debe
dedylja [7]

Answer:

The correct answer is ii) It must revolve around a star

Explanation:

For a celestial body to be called a planet, it must meet at least three characteristics

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* have control over its orbital that is to say to prevent that other body is in its same orbital

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