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A blueberry pancake has diameter 15 cm and contains 6 large blueberries, each of diameter 1 cm. Find the cross section \(\sigma\) of a blueberry and the "target" density \(n_{\text {tar }}\) (number/area) of berries in the pancake, as seen from above. What is the probability that a skewer, jabbed at random into the pancake, will hit a berry (in terms of \(\sigma\) and \(n_{\text {tar }}\) and then numerically)?

Short Answer

Expert verified
The probability is approximately 0.0267 or 2.67%.

Step by step solution

01

Finding the Cross Section of a Blueberry

The cross section of one blueberry is the area of a circle with diameter 1 cm. To find the area, we use the formula for the area of a circle, which is \( \pi r^2 \), where \( r \) is the radius. Since the diameter is 1 cm, the radius \( r \) is 0.5 cm. Calculate the cross-sectional area: \( \sigma = \pi \times (0.5)^2 = \frac{\pi}{4} \text{ cm}^2 \).
02

Calculating the Area of the Pancake

The pancake is also circular, with a diameter of 15 cm, giving it a radius of 7.5 cm. Use the formula for the area of a circle: \( A = \pi r^2 \). Substitute the radius to find the area: \( A = \pi \times (7.5)^2 = 56.25\pi \text{ cm}^2 \).
03

Determining Target Density

Target density \( n_{\text{tar}} \) is defined as the number of blueberries per unit area of pancake. Since there are 6 blueberries, the target density is calculated as: \( n_{\text{tar}} = \frac{6}{56.25\pi} \text{ berries/cm}^2 \approx \frac{6}{176.715} \approx 0.03396 \text{ berries/cm}^2 \).
04

Computing the Probability of Hitting a Berry

To find the probability that a skewer hits a blueberry, we multiply the cross-sectional area \( \sigma \) of a berry by the target density \( n_{\text{tar}} \): \( P = \sigma \times n_{\text{tar}} = \frac{\pi}{4} \times 0.03396 \approx 0.0267 \).
05

Conclusion

The final probability that a skewer jabbed at random hits a berry within the pancake is approximately 0.0267, or 2.67%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

cross section
In geometry, a cross section is the area of an intersection made between a solid object and a plane. Imagine slicing through an object like a pancake. The resulting shape you see would be its cross section. In this exercise, we're interested in the cross section of a blueberry. A blueberry is roughly spherical, so its cross section—when viewed from above, as in this exercise—is a circle.

To find the cross section, we apply the formula for the area of a circle: \( A = \pi r^2 \). Here, \( r \) is the radius of the circle. Since the blueberry’s diameter is 1 cm, its radius is half of this, or 0.5 cm. Therefore, the cross-sectional area \( \sigma \) of a blueberry is calculated as:
  • \( \sigma = \pi \times (0.5)^2 = \frac{\pi}{4} \text{ cm}^2 \)
This gives us the exact area of a typical blueberry's cross section when cut horizontally at its widest diameter.
target density
Target density is a useful measurement, especially in statistics and probability, for determining how densely packed objects are in a given area. It helps in understanding how likely it is for a certain outcome to occur—like hitting a blueberry with a skewer in a pancake.

In our context, target density \( n_{\text{tar}} \) is calculated by counting how many blueberries are spread across the pancake's total area. The pancake, having a diameter of 15 cm, forms a large circle with a calculated area of \( 56.25\pi \text{ cm}^2 \).
  • The number of blueberries is given as 6.
  • Therefore, the target density is \( n_{\text{tar}} = \frac{6}{56.25\pi} \approx 0.03396 \text{ berries/cm}^2 \).
This tells us that, on average, there are approximately 0.034 blueberries on every square centimeter of the pancake. This density plays a crucial role in determining the probability of a skewer hitting a berry.
circle area formula
The area of a circle is one of the most essential geometric formulas. Given a circle’s radius \( r \), the area \( A \) can be found using the formula: \( A = \pi r^2 \). This formula is applied both in finding the cross section of a blueberry and the full area of the pancake in the exercise.

For the blueberry, with a radius of 0.5 cm, the cross-sectional area becomes \( \sigma = \pi \times (0.5)^2 = \frac{\pi}{4} \text{ cm}^2 \). For the pancake, the radius is half of 15 cm, which is 7.5 cm. Thus, the pancake area is:
  • \( A = \pi \times (7.5)^2 = 56.25\pi \text{ cm}^2 \).
Understanding and accurately applying the circle area formula is essential in many math problems involving circular shapes, whether those shapes appear alone or overlap with others in more complex arrangements.

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Most popular questions from this chapter

In their famous experiment, Rutherford's assistants, Geiger and Marsden, detected the scattered alpha particles using a zinc sulphide screen, which produced a tiny flash of light when struck by an alpha particle. If their screen had area \(1 \mathrm{mm}^{2}\) and was \(1 \mathrm{cm}\) from the target, what solid angle did it subtend?

The cross section for scattering a certain nuclear particle by a copper nucleus is 2.0 barns. If \(10^{9}\) of these particles are fired through a copper foil of thickness \(10 \mu \mathrm{m}\), how many particles are scattered? (Copper's density is 8.9 gram/cm \(^{3}\) and its atomic mass is \(63.5 .\) The scattering by any atomic electrons is completely negligible.)

Our definition of the scattering cross section, \(N_{\mathrm{sc}}=N_{\text {inc }} n_{\text {tar }} \sigma,\) applies to an experiment using a narrow beam of projectiles all of which pass through a wide target assembly. Experimenters sometimes use a wide incident beam, which completely engulfs a small target assembly (the beam of photons from a car's headlamp, directed at a small piece of plastic, for example). Show that in this case \(N_{\mathrm{sc}}=n_{\text {inc }} N_{\mathrm{tar}} \sigma\) where \(n_{\text {inc }}\) is the density (number/area) of the incident beam, viewed head-on, and \(N_{\text {tar }}\) is the total number of targets in the target assembly.

Calculate the solid angles subtended by the moon and by the sun, both as seen from the earth. Comment on your answers. (The radii of the moon and sun are \(R_{\mathrm{m}}=1.74 \times 10^{6} \mathrm{m}\) and \(R_{\mathrm{s}}=6.96 \times 10^{8}\) m. Their distances from earth are \(\left.d_{\mathrm{m}}=3.84 \times 10^{8} \mathrm{m} \text { and } d_{\mathrm{s}}=1.50 \times 10^{11} \mathrm{m} .\right)\)

The differential cross section for scattering 6.5-MeV alpha particles at 120^ off a silver nucleus is about 0.5 barns/sr. If a total of \(10^{10}\) alphas impinge on a silver foil of thickness \(1 \mu \mathrm{m}\) and if we detect the scattered particles using a counter of area \(0.1 \mathrm{mm}^{2}\) at \(120^{\circ}\) and \(1 \mathrm{cm}\) from the target, about how many scattered alphas should we expect to count? (Silver has a specific gravity of \(10.5,\) and atomic mass of 108.)

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