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Ainat [17]
3 years ago
11

Ormond Oil company wants to paint one of its cylindrical

Mathematics
1 answer:
Elenna [48]3 years ago
4 0

Answer:

the company must buy 22 gallons to paint this entire area

Step-by-step explanation:

The circumference of the tank is given and is C = 2(pi)r, where r is the area.                                                                              

                                                                                                             118 ft

Here the circumference is C = 2(pi)(r) = 118 ft, which   leads to r = ------------ ≈

18.79 ft ≈ r                                                                                                     2(pi)

The area of the sides is A = (circumference)(height), or approximately

(118 ft)(50 ft) = 5900 ft², and the area of the top is A = πr², which here comes to (π)(18.79 ft)² ≈ 1109 ft².  Combining these two sub-areas, we get:

A(total) = 1109 ft² + 5900 ft² ≈ 7009 ft²

To determine how many gallons of paint will be needed to paint only the top and sides, we divide 7009 ft² by the coverage rate, which is

320 ft²

-----------

1 gallon

which results in:

7009 ft²

----------------------  ≈  21.9 gallons

320 ft² / gallon                          

Since the paint comes only in full gallon cans, the company must buy 22 gallons to paint this entire area.                                                        

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Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

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Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

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3 years ago
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monitta

x = \frac{-11}{3}

<u>Step-by-step explanation:</u>

Here we have , If the first step in the solution of the equation 2x - 8 = 5x + 3 is "subtract 2x," then in the form of a paragraph, explain in complete sentences the next steps necessary to completely solve the equation for x. Let's solve this :

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⇒ - 8 -3= 3x + 3-3 { subtract 3 from both sides }

⇒ - 11 = 3x + 0

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⇒ \frac{-11}{3}  = \frac{3}{3} x { divide 3 from both sides }

⇒ x = \frac{-11}{3}

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Answer:

Step-by-step explanation:

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3 years ago
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I’m think it is the median
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