Question

A Ferris wheel at a carnival has a diameter of meters. Suppose a passenger is traveling at 3 kilometers per hour. (A useful fact: .)
(a) Find the angular speed of the wheel in radians per minute.
(b) Find the number of revolutions the wheel makes per hour. (Assume the wheel does not stop.)

Do not round any intermediate computations, and round your answer to the nearest whole number.

Answer 1

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$\large\underline{\sf{Given :}}$

Radius (r) of the base of the cylinder = 8 cm

Height (h) of the water level = 20 cm

$\large\underline{\sf{To\:find :}}$

Volume of the water in the container = ?

$\large\underline{\sf{Solution :}}$

Formula used :

Volume of cylinder =πr²h

⟶ Volume of the water = (22/7)×8² × 20 cm³

⟶ Volume of the water = (22/7) ×64 ×20 cm³

⟶ Volume of the water = 4022.86 cm³

Therefore,

The volume of the water in the container is 4022.86 cm³✔︎

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★ More info :

• TSA of cylinder = 2πr(h+r)
• CSA of cylinder = 2πrh
• Volume of cone = ⅓ πr²h
• TSA of cone = πr(l+r)
• CSA of cone = πrl

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Answer 2

Answer:

$\huge{\underline{\underline{\bold{\red{\bf{Answer}}}}}}$

Given :-

Side of the square park is 35m

Two semicircle cemented flower beds on two opposite sides of square park

To Find :-

Perimeter of the square park including flower beds

Solution :-

According to the question,

➻ Diameter of the semi circle flower bed is 35m

As we know that,

$\sf{Radius = \dfrac{ Diameter}{2} }$

$\sf{⟼ \: \dfrac{35}{2} }$

➻ First we will find area of the square park :

⠀⠀⠀$\large\boxed{\sf{\pink{Area \: of \: square = ( side ) ²}}}$

By substituting values,

$\sf{⟼(35)^{2} }$

$\sf{⟼1225 {m}^{2} }$

∴ The Area of the Square park is 1225m²

Now,

⠀⠀⠀$\large\boxed{\sf{\pink{Area \: of \: the \: semi \: circle = \dfrac{πr²}{2}}}}$

$\sf{⟼ \dfrac{ \dfrac{22}{7}\times( \dfrac{35}{2})^{2} } {2} }$

$\sf{⟼ \dfrac{ \dfrac{\cancel{22} ^{11}}{\cancel{7} } \times \dfrac{\cancel{35} ^{5} } {\cancel{2} } \times \dfrac{35}{2} }{2} }$

$\sf{⟼ \dfrac{11 \times 5 \times 35}{4} }$

$\sf{⟼ \dfrac{1925}{4} }$

$\sf{⟼481.25 {m}^{2} }$

∴ The Area of the semicircular flower bed is 481.25m²

Now, According to the question,

❶ Area of the square park including flower beds = Area of square park + 2 ( Area of the semicircle flower bed )

By substituting values,

$\sf{⟼1225 + 2(481.25)}$

$\sf{⟼1225 + 962.50}$

$\sf{⟼2187.5 {m}^{2} }$

Hence, The Area of the Square park including semi circular flower beds is 2187.5m²

❷ Perimeter of the square park including flower beds = Perimeter of square park + perimeter of semicircular flower beds

So,

$\sf{⟼2(side) + 2( \dfrac{2\pi r}{2}) }$

By substituting values,

$\sf{⟼2(35) + 2( \dfrac{2 \times \frac{22}{7} \times \frac{35}{2} }{2}) }$

$\sf{⟼(70 )+ 2( \dfrac{2 \times 11 \times 5}{2}) }$

$\sf{⟼70 + 2(11 \times 5)}$

$\sf{⟼70 +2(55)}$

$\sf{⟼70 + 110}$

$\sf{⟼180m}$

Hence,

The Area of the square park including semicircular flower beds is 180m