Answer/Step-by-step explanation:
Problem 1:
Radius = 4.8 cm
Height = 6 cm
Volume of cylinder (V) = πr²h
Plug in the values
V = π*4.8²*6 = 434.29 cm³
Problem 2:
Length of pipe = 26 cm
Internal diameter = 6.5 cm
Thickness = 0.5 cm
Pipe volume (V) = π(R² - r²)h
where,
R = Outer radius = ½(6.5) + 0.5 = 3.75 cm
r = inner radius = ½(6.5) = 3.25 cm
h = height = 26 cm
Plug in the values
V = π(3.75² - 3.25²)*26 = 285.88 cm³
Problem 3:
Volume the cylindrical paint can hold = 2.5 litres = 2.5*1000 = 2,500 cm³
Height (h) = 16 cm
Radius (r) = ??
Volume of cylindrical can (V) = πr²h
Plug in the values
2,500 = π × r² × 16
2,500 = 16π × r²
Divide both sides by 16π
2500/16π = r²
49.7 = r²
Take the square root of both sides
√49.7 = r
r = 7.05 cm (nearest hundredth)
Problem 4:
The section of the guttering is ½ of a cylinder
Diameter = 14 cm = 0.14 m
Radius = ½(0.14) = 0.07 m
Volume = 20 litres = 0.02 m³
Length (h) = ??
Volume of the guttering = ½(volume of cylinder) = ½(πr²h)
Plug in the values
0.02 = ½(π*0.07²*h)
0.02*2 = 0.0049π*h
0.04 = 0.0049π*h
Divide both sides by 0.0049π
0.04/0.0049π = h
2.6 = h (nearest tenth)
Length = 2.6 m
Problem 5:
Height of the smaller cylinder (h) = 13 cm
Radius of the smaller cylinder (r) = ½(7) = 3.5 cm
Volume of smaller cylinder = πr²h = π × 3.5² × 13 = 500.3 cm³
Volume of larger cylinder filled to a height of 5 cm = Volume of smaller cylinder
Thus:
Volume of cylinder filled to height of 5cm = 500.3 cm³
height (h) = 5
radius (r) = ???
Therefore,
500.3 = π × r² × 5
500.3 = 5π × r²
500.3/5π = r²
31.9 = r²
√31.9 = r
r = 5.6 cm (nearest tenth)