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A cube \(5.0 \mathrm{~cm}\) on each side is made of a metal alloy. After you drill a cylindrical hole \(2.0 \mathrm{~cm}\) in diameter all the way through and perpendicular to one face, you find that the cube weighs \(6.30 \mathrm{~N}\). (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?

Short Answer

Expert verified
(a) The density of the metal is 5.88 g/cm^3. (b) The cube weighed 7.203N before the hole was drilled.

Step by step solution

01

Calculate the total volume of the cube

Given that the length of each side of the cube is 5.0 cm, the total volume of the cube can be calculated using the formula for the volume of a cube, which is \(side^3\). So the total volume is \(5.0 cm * 5.0 cm * 5.0 cm = 125.0 cm^3.\)
02

Calculate the volume of the cylindrical hole

The cylindrical hole has a diameter of 2.0 cm, which means the radius will be \(1.0 cm\). Hence, the volume of the cylindrical hole can be found by using the formula for the volume of a cylinder, which is \(\pi * radius^2 * height\). Here, the height of the cylinder is the same as the side of the cube which is 5cm. So, the volume of the hole is \(3.14 * 1.0 cm * 1.0 cm * 5.0 cm = 15.7 cm^3.\)
03

Calculate the volume of the drilled cube

The volume of the drilled cube is obtained by subtracting the volume of the cylindrical hole from the total volume of the cube i.e. \(125.0 cm^3 - 15.7 cm^3 = 109.3 cm^3.\)
04

Calculate the density of the drilled cube

Using the density formula, which is \(density = mass/volume\), and knowing that the weight of the cube is 6.30N (where \(1N = 0.102 kg\) due to gravity), the mass of the cube is \(6.3N * 0.102 kg/N = 0.6426 kg\). Hence the density of the cube is \(0.6426kg / 109.3 cm^3 = 0.00588 kg/cm^3 = 5.88 g/cm^3.\)
05

Calculate the original weight of the cube

The weight of the cube originally before the hole was drilled can be calculated using the density formula rearranged to solve for mass \(mass = density * volume\) and the weight formula \(weight = mass * gravity\). So, mass is \(5.88 g/cm^3 * 125.0 cm^3 = 735.0 g = 0.735 kg\), and weight is \(0.735 kg * 9.8N/kg = 7.203N.\)

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Key Concepts

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

Volume of a Cube
Understanding the volume of a cube is fundamental in mathematics and physics as it's one of the simplest forms to calculate. The volume of a cube can be found by multiplying the length of one of its sides by itself three times. This is expressed mathematically as V = s^3, where V represents volume and s is the length of a side. For instance, if each side of a cube measures 5.0 cm, then the volume is 5.0 cm x 5.0 cm x 5.0 cm = 125.0 cm^3.

Remember, the unit of volume is always a cubic measurement, whether it’s cubic centimeters (cm³), cubic meters (m³), or another cubic unit. This measurement tells us the amount of space inside the three-dimensional object. The concept is crucial when determining the capacity of objects or, in the case of physics, when you are working with density calculations.
Volume of a Cylinder
The volume of a cylinder is calculated differently than a cube due to its circular base. Here, you need to know the radius (r) – which is half the cylinder's diameter – and the height (h). The formula for a cylinder's volume is V = Ï€°ù²h. Let's break it down: The term Ï€°ù² gives the area of the circular base, and when you multiply it by the height, you obtain the total volume.

In our exercise, a cylindrical hole is drilled through the cube, so we must know how to find this volume to subtract it from the cube’s total volume. With a cylinder of radius 1.0 cm and height 5.0 cm, the volume is 3.14 x (1.0 cm)² x 5.0 cm = 15.7 cm³. It’s vital to work with the correct units and to be precise with numbers like π to ensure accurate calculations.
Mass and Weight Conversion
Mass and weight are often used interchangeably in daily conversation, but in physics they have different meanings. Mass refers to the amount of matter in an object and is measured in kilograms (kg), grams (g), and so forth. Weight, on the other hand, is the force exerted by gravity on that mass and is measured in newtons (N).

To convert weight to mass, you can use the relationship weight = mass x gravity, with Earth’s gravity approximately equal to 9.8 m/s². Therefore, if an object weighs 6.30N, its mass is calculated by 6.30N / 9.8N/kg = 0.6426 kg. To solve physics problems, switching between mass and weight is common, and understanding how to perform these conversions accurately is crucial for problems related to density and force.

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Most popular questions from this chapter

At a certain point in a horizontal pipeline, the water's speed is \(2.50 \mathrm{~m} / \mathrm{s}\) and the gauge pressure is \(1.80 \times 10^{4} \mathrm{~Pa}\). Find the gauge pressure at a second point in the line if the cross-sectional area at the second point is twice that at the first.

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A cubical block of density \(\rho_{\mathrm{B}}\) and with sides of length \(L\) floats in a liquid of greater density \(\rho_{\mathrm{L}}\). (a) What fraction of the block's volume is above the surface of the liquid? (b) The liquid is denser than water (density \(\rho_{\mathrm{W}}\) ) and does not mix with it. If water is poured on the surface of that liquid, how deep must the water layer be so that the water surface just rises to the top of the block? Express your answer in terms of \(L, \rho_{\mathrm{B}}, \rho_{\mathrm{L}},\) and \(\rho_{\mathrm{W}}\). (c) Find the depth of the water layer in part (b) if the liquid is mercury, the block is made of iron, and \(L=10.0 \mathrm{~cm}\).

Water runs into a fountain, filling all the pipes, at a steady rate of \(0.750 \mathrm{~m}^{3} / \mathrm{s}\). (a) How fast will it shoot out of a hole \(4.50 \mathrm{~cm}\) in diameter? (b) At what speed will it shoot out if the diameter of the hole is three times as large?

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 2200.355 -L cans per minute. At point 2 in the pipe, the gauge pressure is \(152 \mathrm{kPa}\) and the cross-sectional area is \(8.00 \mathrm{~cm}^{2}\). At point \(1,1.35 \mathrm{~m}\) above point \(2,\) the cross- sectional area is \(2.00 \mathrm{~cm}^{2}\). Find the (a) mass flow rate; (b) volume flow rate; (c) flow speeds at points 1 and \(2 ;\) (d) gauge pressure at point 1 .

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