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A solid aluminum ingot weighs 89 N in air. (a) What is its volume? (b) The ingot is suspended from a rope and totally immersed in water. What is the tension in the rope (the \(apparent\) weight of the ingot in water)?

Short Answer

Expert verified
(a) The volume is \(0.00336 \text{ m}^3\); (b) The apparent weight is \(56.03 \text{ N}\).

Step by step solution

01

Understand the given data

The solid aluminum ingot weighs 89 N in air. We are required to find the volume of the ingot and the apparent weight when it is submerged in water. Aluminum's density \( \rho_{al} \) is approximately 2700 kg/m³.
02

Calculate the Volume of the Ingot

The weight of the object is given by \( W = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \)).To find the volume, use the relationship \( m = \rho \cdot V \), where \( \rho \) is the density and \( V \) is the volume.We start by solving for the mass: \[m = \frac{W}{g} = \frac{89}{9.81} \approx 9.07 \text{ kg}\]Then, we use the density formula:\[V = \frac{m}{\rho_{al}} = \frac{9.07}{2700} \approx 0.00336 \text{ m}^3\]
03

Calculate the Apparent Weight in Water

The apparent weight in water is the actual weight minus the buoyant force. The buoyant force is equal to the weight of the water displaced by the ingot.First, calculate the buoyant force:\[F_{buoyant} = \text{density of water} \times V \times g = 1000 \times 0.00336 \times 9.81 \approx 32.97 \text{ N}\]Then, calculate the apparent weight:\[W_{apparent} = W - F_{buoyant} = 89 - 32.97 \approx 56.03 \text{ N}\]

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Key Concepts

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

Apparent Weight
When an object is submerged in a fluid, it experiences a reduction in weight known as the apparent weight. This phenomenon is due to the buoyant force that acts upward against gravity. The buoyant force makes objects feel lighter when they are in a liquid, like water.
In the exercise, the apparent weight of the aluminum ingot is the weight it seems to have when submerged in water. To find this, you subtract the buoyant force from the actual weight of the ingot.
  • The actual weight in air is 89 N.
  • The buoyant force equals the weight of the water displaced by the ingot.
  • The formula for buoyant force is: \[ F_{buoyant} = \text{density of water} \times V \times g \]
Subsequently, the apparent weight can be calculated as: \[ W_{apparent} = W - F_{buoyant} \]In our case, the apparent weight is approximately 56.03 N after subtracting the buoyant force of 32.97 N from the initial weight of 89 N.
Density
Density is a measure of how much mass is contained in a given volume. It's an essential concept in physics, especially when dealing with buoyancy and fluid mechanics.
The density is defined by the formula: \( \rho = \frac{m}{V} \), where \( \rho \) is density, \( m \) is mass, and \( V \) is volume.
The density of aluminum is approximately 2700 kg/m³ which is used to calculate the mass of the ingot in the problem. Since the density of water is 1000 kg/m³, we can use this property to predict how much fluid is displaced by the ingot.
  • Higher density means more mass in each unit of volume.
  • Objects denser than the fluid will sink; less dense objects will float.
This principle helps in determining the buoyant force, which impacts the apparent weight of the object in water. With these insights, students can correctly apply density in related physics problems.
Volume Calculation
The calculation of volume is crucial to understanding the use of density in physics. To find the volume of an object when the mass and the density are given, you can rearrange the density formula to:
\[ V = \frac{m}{\rho} \]This formula allows you to find the space an object occupies.
In the exercise about the aluminum ingot, once the mass is calculated (using weight divided by gravity), you can then substitute the mass and aluminum density values into this formula to find the volume.
  • First, determine the mass using: \( m = \frac{W}{g} \).
  • Next, calculate the volume with: \( V = \frac{9.07}{2700} \approx 0.00336 \text{ m}^3 \).
By finding the volume, students can apply it to determine the amount of fluid displaced by the object when submerged, thus calculating buoyant force and apparent weight. Volume calculation is thereby foundational for solving various problems in fluid mechanics.

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