/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 A uniform lead sphere and a unif... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform lead sphere and a uniform aluminum sphere have the same mass. What is the ratio of the radius of the aluminum sphere to the radius of the lead sphere?

Short Answer

Expert verified
The ratio of the radius of the aluminum sphere to the radius of the lead sphere is approximately 1.53.

Step by step solution

01

Recall the formula for the volume of a sphere.

The formula for the volume of a sphere is given by \( V = \frac{4}{3} \pi \ r^{3} \) where \( r \) is the radius.
02

Express the mass of a sphere.

The mass of a sphere is given by \( m = \rho V \), where \( \rho \) is the density of the material and \( V \) is the volume of the sphere. Therefore, using the definition of \( V \), we have \( m = \rho \frac{4}{3} \pi \ r^{3} \).
03

Set up and solve the ratio.

Given that the masses of the two spheres are equal, the ratio of the radii cubed of the aluminum sphere (\( r_{A} \)) to the lead sphere (\( r_{L} \)) is equal to the inverse ratio of their densities:\( \frac{r_{A}^{3}}{r_{L}^{3}} = \frac{\rho_{L}}{\rho_{A}} \). To find \( \frac{r_{A}}{r_{L}} \), we take the cube root of \( \frac{\rho_{L}}{\rho_{A}} \). Using known values for the densities of lead (11340 kg/m³) and aluminium(2700 kg/m³), we get \( \frac{r_{A}}{r_{L}} = \sqrt[3]{\frac{11340}{2700}} \).

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Key Concepts

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

Sphere Volume
To understand the concept of sphere volume, imagine inflating a balloon. As it gets larger, the volume inside increases. Mathematically, the volume \( V \) of a sphere is calculated using the formula:
  • \( V = \frac{4}{3} \pi r^3 \)
Here, \( r \) represents the radius of the sphere, which is the distance from its center to any point on its surface. The constant \( \pi \) is approximately 3.14159, though it's used in its exact form in calculations when possible.
The cubed radius \( r^3 \) signifies that the volume grows immensely with even small increases in radius, as the radius is multiplied by itself twice. Therefore, if you double the radius, the volume increases by a factor of eight \((2^3 = 8)\). Remember, the volume captured by a sphere depends heavily on its radius, a core principle that applies to our exercise problem.
Density Relation
Density is a measure of how much mass occupies a particular volume. In simpler terms, it's essentially how "heavy" something feels for its size. This is calculated with the formula:
  • \( \rho = \frac{m}{V} \)
Here, \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume.
In the given exercise problem, the concept of density is essential as it relates directly to both materials' properties: lead and aluminum. Although both spheres have the same mass, their densities are different, impacting their respective volumes, and consequently, their radii. Since lead is denser than aluminum, an equivalent mass of lead will occupy less volume than the same mass of aluminum.
This relationship explains why in our solution, the ratio of the radii is inversely proportional to the ratio of the densities. Understanding density helps us grasp why lighter materials take up more space and heavier ones take up less.
Cube Root Calculation
Cube root calculations are about finding a number that, when multiplied by itself three times, equals another number. It's like reversing the process of cubing a number.
In the context of the problem, we need to find the cube root to determine the ratio of radii given the radii are related by their cubed values. Let's see this relationship mathematically:
  • \( \frac{r_A^3}{r_L^3} = \frac{\rho_L}{\rho_A} \)
  • \( \frac{r_A}{r_L} = \sqrt[3]{\frac{\rho_L}{\rho_A}} \)
In our problem, the cube root is taken of the density ratio \( \frac{11340}{2700} \), simplifying the calculation to find the radius ratio. The cube root is a necessary step here because the initial equation gives us a relation in terms of cube powers, and finding the cube root simplifies that relation to direct radii. Remember, the cube root calculation translates abstract mathematical ratios into something tangible like the radius, accommodating easier understanding of practical problems.

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Most popular questions from this chapter

A plastic ball has radius \(12.0 \mathrm{~cm}\) and floats in water with \(24.0 \%\) of its volume submerged. (a) What force must you apply to the ball to hold it at rest totally below the surface of the water? (b) If you let go of the ball, what is its acceleration the instant you release it?

A liquid flowing from a vertical pipe has a definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed \(v_{0}\) and the radius of the stream of liquid is \(r_{0}\). (a) Find an equation for the speed of the liquid as a function of the distance \(y\) it has fallen. Combining this with the equation of continuity, find an expression for the radius of the stream as a function of \(y\). (b) If water flows out of a vertical pipe at a speed of \(1.20 \mathrm{~m} / \mathrm{s},\) how far below the outlet will the radius be one-half the original radius of the stream?

How does the force the diaphragm experiences due to the difference in pressure between the lungs and abdomen depend on the abdomen's distance below the water surface? The force (a) increases linearly with distance; (b) increases as distance squared; (c) increases as distance cubed; (d) increases exponentially with distance.

The Environmental Protection Agency is investigating an abandoned chemical plant. A large, closed cylindrical tank contains an unknown liquid. You must determine the liquid's density and the height of the liquid in the tank (the vertical distance from the surface of the liquid to the bottom of the tank). To maintain various values of the gauge pressure in the air that is above the liquid in the tank, you can use compressed air. You make a small hole at the bottom of the side of the tank, which is on a concrete platform - so the hole is \(50.0 \mathrm{~cm}\) above the ground. The table gives your measurements of the horizontal distance \(R\) that the initially horizontal stream of liquid pouring out of the tank travels before it strikes the ground and the gauge pressure \(p_{\mathrm{g}}\) of the air in the tank. $$\begin{array}{l|lllll}p_{\mathrm{g}}(\mathrm{atm}) & 0.50 & 1.00 & 2.00 & 3.00 & 4.00 \\\\\hline \boldsymbol{R}(\mathrm{m}) & 5.4 & 6.5 & 8.2 & 9.7 & 10.9\end{array}$$ (a) Graph \(R^{2}\) as a function of \(p_{\mathrm{g}}\). Explain why the data points fall close to a straight line. Find the slope and intercept of that line. (b) Use the slope and intercept found in part (a) to calculate the height \(h\) (in meters) of the liquid in the tank and the density of the liquid (in \(\mathrm{kg} / \mathrm{m}^{3}\) ). Use \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} .\) Assume that the liquid is nonviscous and that the hole is small enough compared to the tank's diameter so that the change in \(h\) during the measurements is very small.

A gallon of milk in a full plastic jug is sitting on the edge of your kitchen table. Estimate the vertical distance between the top surface of the milk and the bottom of the jug. Also estimate the distance from the tabletop to the floor. You punch a small hole in the side of the jug just above the bottom of the jug, and milk flows out the hole. When the milk first starts to flow out the hole, what horizontal distance does it travel before reaching the floor? Assume the milk is in free fall after it has passed through the hole, and neglect the viscosity of the milk.

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