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What is the terminal speed of a \(6.00 \mathrm{~kg}\) spherical ball that has a radius of \(3.00 \mathrm{~cm}\) and a drag coefficient of \(1.60 ?\) The density of the air through which the ball falls is \(1.20 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The terminal speed of the ball is approximately 147.16 m/s.

Step by step solution

01

Calculate the Gravitational Force

The gravitational force is the weight of the ball. Use the formula: \[ F_g = m \cdot g \] where \( m = 6.00 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \). \[ F_g = 6.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 58.86 \, \text{N} \]
02

Determine the Drag Force at Terminal Velocity

At terminal velocity, the drag force equals the gravitational force. The drag force is given by: \[ F_d = \frac{1}{2} \cdot C_d \cdot \rho \cdot A \cdot v_t^2 \] where \( C_d = 1.60 \), \( \rho = 1.20 \, \text{kg/m}^3 \) and \( A \) is the cross-sectional area. To find \( A \), use:\[ A = \pi \cdot r^2 \]. First, convert radius from cm to m: \( r = 3.00 \, \text{cm} = 0.03 \, \text{m} \) Then calculate area: \[ A = \pi \times (0.03)^2 = 2.827 \times 10^{-3} \, \text{m}^2 \].
03

Equate Forces to Solve for Terminal Velocity

Since \( F_g = F_d \) at terminal velocity: \[ 58.86 = \frac{1}{2} \times 1.60 \times 1.20 \times 2.827 \times 10^{-3} \times v_t^2 \] Simplify and solve for \( v_t^2 \): \[ v_t^2 = \frac{58.86}{0.5 \times 1.60 \times 1.20 \times 2.827 \times 10^{-3}} \].
04

Calculate Terminal Velocity

Plug in the numbers and calculate: \[ v_t^2 = \frac{58.86}{0.0027168} = 21668.3 \] Now, take the square root to find \( v_t \): \[ v_t = \sqrt{21668.3} = 147.16 \, \text{m/s} \].

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Key Concepts

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

Gravitational Force
The gravitational force, often referred to as weight, is the force exerted by the Earth on any object with mass. This force pulls objects towards the center of the Earth. In this exercise, the gravitational force is calculated using the formula: \( F_g = m \times g \), where \( m \) denotes the mass of the spherical ball and \( g \) represents the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). For a 6.00 kg ball, the gravitational force amounts to \( 58.86 \, \text{N} \). This constant force acts throughout the descent of the ball, attempting to accelerate it downward. Understanding this force is key to balancing it with the upward drag force as the ball reaches terminal velocity.
Drag Force
The drag force acts in opposition to an object's motion through a fluid, such as air. As objects fall, they face resistance from the air, which slows them down. This force can be calculated using the formula: \( F_d = \frac{1}{2} \times C_d \times \rho \times A \times v_t^2 \), where:
  • \( C_d \) is the drag coefficient, indicating how "streamlined" an object is.
  • \( \rho \) is the air density.
  • \( A \) is the cross-sectional area of the object.
  • \( v_t \) is the terminal velocity.
At terminal velocity, this drag force equals the gravitational force, meaning the net force is zero and the object falls at a constant speed.
Drag Coefficient
The drag coefficient, \( C_d \), is a dimensionless number characterizing an object's resistance to motion through a fluid such as air. A lower drag coefficient indicates that an object is more streamlined and experiences less drag. In this problem, the drag coefficient is given as 1.60. This value helps in calculating the drag force acting on the ball as it moves through the air. It's determined by factors such as the shape of the object and the flow conditions around it. A sphere might typically have a drag coefficient ranging from 0.47 to 1.0, but it can be higher, depending on surface textures and other conditions. Understanding \( C_d \) is vital in predicting how quickly or slowly an object might reach its terminal velocity.
Air Density
Air density, represented by \( \rho \), impacts how objects move through the air. It measures the mass of air per unit volume, typically expressed in kg/m³. In our exercise, the air density is given as 1.20 kg/m³. Air density can vary due to changes in altitude, temperature, and humidity. Denser air creates more drag force on a moving object, slowing it down more compared to less dense air. High air density leads to increased resistance and is a key factor in calculating the drag force, as seen in the formula \( F_d = \frac{1}{2} \times C_d \times \rho \times A \times v_t^2 \). By understanding this concept, you can better predict how quickly an object might reach its terminal velocity in varying environments.

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