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Aleks04 [339]
2 years ago
5

A spherical balloon is filled with gas at a rate of 4 cm 3 /s . at what rate is the radius r changing with with respect to time

when the volume v = 36 π cm 3 ?
Physics
1 answer:
Gnoma [55]2 years ago
8 0
The volume of the sphere is given by V = 4/3 * pie * r^3. We seek dr/dt that the rate of change of the radius with respect to time.  
V = 4/3 * pie * r^3 
Since we know the rate at which the volume changes wrt time. We can plug it in to find dr/dt.
 dV/dt = 36pie. But dV/dt = 4/3 * pie * 3r^2 * dr/dt
 So we have that dr/dt = (dV/dt) / 4/3 * pie * 3r^2
 So dr/dt = (36pie)/ 4/3 * pie * 3(4)^2
 dr/dt = 113.112/ 201.0864 = 0.5624

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The concentration of a solution that contains 70g of H2SO4 in 0,28 dm³ of solution is?​
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Taking into account the definition of molarity, the concentration of a solution that contains 70 g of H₂SO₄ in 0,28 dm³ of solution is 2.5510 \frac{moles}{L}.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

Molarity=\frac{number of moles}{volume}

Molarity is expressed in units \frac{moles}{L}.

<h3>This case</h3>

In this case, you have:

  • number of moles= 70 g×\frac{1 mole}{98 g}= 0.7143 moles, where 98 g/mole os the molar mass of H₂SO₄
  • volume= 0.28 dm³= 0.28 L (being 1 dm³= 1 L)

Replacing in the definition of molarity:

Molarity=\frac{0.7143 moles}{0.28 L}

Solving:

<u><em>Molarity= 2.5510 </em></u>\frac{moles}{L}

Finally, the concentration of a solution that contains 70 g of H₂SO₄ in 0,28 dm³ of solution is 2.5510 \frac{moles}{L}.

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A chocolate chip cookie is an example of a (2 points) a homogeneous mixture b heterogeneous mixture c suspension d colloid
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I think it is heterogeneous mixture. have a good day

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The speed of sound on planet is 210 m/s.

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Oscillation is the repeating or periodic change of a quantity around a central value or between two or more states, often in time. Alternating current and a swinging pendulum are two common examples of oscillation.

There are 3 main types of Oscillation –

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f = frequency = 600 Hz

lambda = wavelength = 35 cm = 0.35 m

Now,

V = speed = f × lambda = 210 m/s

Hence, speed of sound on planet is 210 m/s.

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Consider a semicircular ring of radius R. Its linear mass density varies as lambda =lambda not sin theta. Locate its centre of m
bearhunter [10]

Answer:

(0, πR/4)

Explanation:

The linear mass density (mass per length) is λ = λ₀ sin θ.

A short segment of arc length is ds = R dθ.

The mass of this short length is:

dm = λ ds

dm = (λ₀ sin θ) (R dθ)

dm = R λ₀ sin θ dθ

The x coordinate of the center of mass is:

X = ∫ x dm / ∫ dm

X = ∫₀ᵖ (R cos θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

X = R ∫₀ᵖ sin θ cos θ dθ / ∫₀ᵖ sin θ dθ

X = R ∫₀ᵖ ½ sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R ∫₀ᵖ 2 sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R (-cos 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

X = ¼R (-cos 2π − (-cos 0)) / (-cos π − (-cos 0))

X = ¼R (-1 + 1) / (1 + 1)

X = 0

The y coordinate of the center of mass is:

Y = ∫ y dm / ∫ dm

Y = ∫₀ᵖ (R sin θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

Y = R ∫₀ᵖ sin² θ dθ / ∫₀ᵖ sin θ dθ

Y = R ∫₀ᵖ ½ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R ∫₀ᵖ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R (θ − ½ sin 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

Y = ½R [(π − ½ sin 2π) − (0 − ½ sin 0)] / (-cos π − (-cos 0))

Y = ½R (π − 0) / (1 + 1)

Y = ¼πR

4 0
3 years ago
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