Problem 89
Question
Thickness of an Oil Film (Refer to Example 9.) A drop of oil measuring 0.12 cubic centimeter is spilled onto a lake. The oil spreads out in a circular shape having a diameter of 23 centimeters. Approximate the thickness of the oil film.
Step-by-Step Solution
Verified Answer
The thickness of the oil film is approximately 0.000289 cm.
1Step 1: Calculate the Area of the Circular Oil Spill
The oil spreads in a circular shape. The formula for the area of a circle is \( A = \pi r^2 \), where \( r \) is the radius. Since the diameter is 23 cm, the radius \( r \) is \( \frac{23}{2} = 11.5 \) cm. Plug \( 11.5 \) cm into the area formula: \( A = \pi \times (11.5)^2 \approx 415.26 \) cm².
2Step 2: Calculate the Thickness of the Oil Film
The volume of the oil is given as 0.12 cm³. To find the thickness (\( t \)) of the oil film, use the formula \( t = \frac{V}{A} \), where \( V \) is the volume and \( A \) is the area. So, \( t = \frac{0.12}{415.26} \approx 0.000289 \) cm.
Key Concepts
Volume calculationArea of a circleThickness calculation
Volume calculation
Volume calculation is an essential part of solving problems related to three-dimensional objects. In our scenario, the volume is given. But generally speaking, to find the volume of simple shapes, you use specific formulas depending on the shape. For example:
\[ \text{Volume of a cube} = \text{side}^3 \]
\[ \text{Volume of a cylinder} = \pi r^2 h \]
Here, the volume we are working with is 0.12 cubic centimeters, which represents the amount of oil spilled. This measurement is crucial as it helps determine other properties, like the thickness of the oil film, when combined with additional measurements like the area.
\[ \text{Volume of a cube} = \text{side}^3 \]
\[ \text{Volume of a cylinder} = \pi r^2 h \]
Here, the volume we are working with is 0.12 cubic centimeters, which represents the amount of oil spilled. This measurement is crucial as it helps determine other properties, like the thickness of the oil film, when combined with additional measurements like the area.
Area of a circle
The area of a circle is found using the formula \( A = \pi r^2 \), which signifies the space enclosed within a circle's boundary. "\( \pi \)" is a mathematical constant approximately equal to 3.14159. The variable \( r \) stands for the radius, which is half the circle's diameter.
In the context of the oil spill, the circle formed by the oil has a diameter of 23 centimeters. To find the radius, simply divide the diameter by 2, giving us 11.5 centimeters. Once you have the radius, plug it into the area formula:
\[ A = \pi (11.5)^2 \approx 415.26 \text{ cm}^2 \]
This result provides the size of the surface area that the oil covers on the water.
In the context of the oil spill, the circle formed by the oil has a diameter of 23 centimeters. To find the radius, simply divide the diameter by 2, giving us 11.5 centimeters. Once you have the radius, plug it into the area formula:
\[ A = \pi (11.5)^2 \approx 415.26 \text{ cm}^2 \]
This result provides the size of the surface area that the oil covers on the water.
Thickness calculation
The thickness of an object, in this case, an oil film, is essentially its depth or height when laid flat. To find the thickness \( t \) of the oil film, we use the formula:
\[ t = \frac{V}{A} \]
where \( V \) is the volume of the oil and \( A \) is the area of the oil spread. This formula is derived from considering that volume is evenly distributed over an area.
With the oil's volume given as 0.12 cm³ and its area calculated as approximately 415.26 cm², the thickness becomes:
\[ t = \frac{0.12}{415.26} \approx 0.000289 \text{ cm} \]
This calculation tells us how thinly the oil is spread on the surface of the water, highlighting the delicate nature of the spill on the lake.
\[ t = \frac{V}{A} \]
where \( V \) is the volume of the oil and \( A \) is the area of the oil spread. This formula is derived from considering that volume is evenly distributed over an area.
With the oil's volume given as 0.12 cm³ and its area calculated as approximately 415.26 cm², the thickness becomes:
\[ t = \frac{0.12}{415.26} \approx 0.000289 \text{ cm} \]
This calculation tells us how thinly the oil is spread on the surface of the water, highlighting the delicate nature of the spill on the lake.
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