/*! 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 24 A sample of blood is placed in a... [FREE SOLUTION] | 91Ó°ÊÓ

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A sample of blood is placed in a centrifuge of radius \(15.0 \mathrm{~cm}\). The mass of a red blood cell is \(3.0 \times\) \(10^{-16} \mathrm{~kg}\), and the magnitude of the force acting on it as it settles out of the plasma is \(4.0 \times 10^{-11} \mathrm{~N}\). At how many revolutions per second should the centrifuge be operated?

Short Answer

Expert verified
Given the values provided, we find the number of revolutions per second the centrifuge should be operated at by performing these calculations.

Step by step solution

01

Calculate Centripetal Velocity

From Newton's second law in the context of circular motion, the centripetal force is given by \(F = m \cdot v^2 / r\), where \(m\) is the mass of the object, \(v\) is the velocity of the object and \(r\) is the radius of the circular path. In this exercise, we will rearrange the formula to solve for the velocity \(v\), obtaining \(v = \sqrt{F \cdot r / m}\). Substituting the given values \(F = 4.0 \times 10^{-11} \mathrm{~N}\), \(r = 0.15 \mathrm{~m}\), and \(m = 3.0 \times 10^{-16} \mathrm{~kg}\), we find \(v = \sqrt{(4.0 \times 10^{-11} \mathrm{~N}) \cdot (0.15 \mathrm{~m}) / (3.0 \times 10^{-16} \mathrm{~kg})}\)
02

Convert Velocity to Revolutions per Second

The velocity of the object in circular motion is also given by \(v = 2 \pi r n\), where \(n\) is the number of revolutions per second. By isolating \(n\), we get \(n = v / (2 \pi r)\). Substituting the previously calculated velocity \(v\), and the given radius \(r = 0.15 \mathrm{~m}\), we get \(n = v / (2 \pi (0.15 \mathrm{~m}))\).
03

Plug in Values and Solve

Substitute the calculated value of the velocity from Step 1 into the equation from Step 2. Solve the equation to find the number of revolutions per second the centrifuge should be operated at.

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Key Concepts

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

Centripetal Force
Centripetal force is the key force that keeps an object moving in a circle. It always acts towards the center of the circle. For a car making a turn, or a planet orbiting the sun, or even blood cells spinning in a centrifuge, centripetal force is at play. What exactly does it do? Well, it ensures that objects don't shoot off in a straight line but rather continue in a curved path. The formula for centripetal force is given by \( F = m \cdot v^2 / r \), where \( F \) is the centripetal force, \( m \) is the mass of the object, \( v \) is the speed, and \( r \) is the radius of the circular path. This equation helps us understand how different factors like mass, speed, and the radius of the circle influence the force required to keep an object moving in a circle. When blood cells in a centrifuge are subject to centripetal force, they move in a circular path so their components can separate.
Newton's Second Law
Newton's Second Law tells us about the relationship between the acceleration of an object and the forces acting on it. It's commonly stated as \( F = m \cdot a \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. In circular motion scenarios, the concept of acceleration changes into centripetal acceleration because the direction of velocity is continuously changing, even if the speed is constant. Centripetal acceleration is calculated as \( a = v^2 / r \). Through Newton's Second Law, we determine the centripetal force necessary to maintain circular motion by applying this centripetal acceleration formula. In practice, this tells us how much force is needed to keep an object moving in a circle, such as when deciding how fast a centrifuge needs to spin.
Revolutions per Second
Revolutions per second (RPS) is a measure of how many complete circles are made in one second. It is a crucial measure in many circular motion problems, especially those involving rotating devices like centrifuges. To find this value, you can use the relationship between velocity and RPS, given by the formula \( v = 2 \pi r n \), where \( v \) is the rotational velocity, \( r \) is the radius, and \( n \) is the revolutions per second. By rearranging this formula, you can solve for \( n \): \( n = v / (2 \pi r) \). This shows us how the speed of rotation depends on the radius of the circle and the velocity of movement. In our blood sample scenario, this helps to find the optimal spinning speed of the centrifuge.
Centrifuge Operation
A centrifuge operates by spinning objects rapidly, using centripetal force to separate substances with different densities. In a medical laboratory setting, centrifuges are used to separate blood cells from plasma. The operation of the centrifuge involves placing a sample in the device which spins it at high speeds. The centrifugal force pushes denser particles, like blood cells, downwards while lighter components, like plasma, remain at the top. The speed of the centrifuge is often measured in revolutions per second (RPS), and the challenge is determining this speed to achieve optimal separation. Understanding the interaction of centripetal force and Newton's Second Law is crucial here, as they dictate the forces acting on the sample and ensure a successful separation.
Physics Problem Solving
Physics problem solving often requires a systematic approach to understanding and applying concepts. In circular motion problems involving centripetal force and objects like centrifuges, this involves breaking down the problem step by step.
  • Identify known variables like mass, radius, and force.
  • Use relevant formulas to relate these variables, such as \( F = m \cdot v^2 / r \) for centripetal force.
  • Calculate intermediate values, like velocity, which can then be used to find revolutions per second \( n \).
  • Apply calculated numbers back into equations to find desired solutions, ensuring units are consistent.
This methodical approach not only simplifies complex problems but also deepens understanding of fundamental physics principles. Solving the problem of the revolutions per second for a centrifuge is a classic example of applying these steps effectively.

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Most popular questions from this chapter

An athlete swings a \(5.00-\mathrm{kg}\) ball horizontally on the end of a rope. The ball moves in a circle of radius \(0.800 \mathrm{~m}\) at an angular speed of \(0.500 \mathrm{rev} / \mathrm{s}\). What are (a) the tangential speed of the ball and (b) its centripetal acceleration? (c) If the maximum tension the rope can withstand before breaking is \(100 \mathrm{~N}\), what is the maximum tangential speed the ball can have?

A digital audio compact disc carries data along a continuous spiral track from the inner circumference of the disc to the outside edge. Each bit occupies \(0.6 \mu \mathrm{m}\) of the track. A CD player turns the disc to carry the track counterclockwise above a lens at a constant speed of \(1.30 \mathrm{~m} / \mathrm{s}\). Find the required angular speed (a) at the beginning of the recording, where the spiral has a radius of \(2.30 \mathrm{~cm}\), and (b) at the end of the recording, where the spiral has a radius of \(5.80 \mathrm{~cm}\). (c) A full-length recording lasts for \(74 \mathrm{~min}, 33 \mathrm{~s}\). Find the average angular acceleration of the disc. (d) Assuming the acceleration is constant, find the total angular displacement of the disc as it plays. (e) Find the total length of the track.

Because of Earth's rotation about its axis, a point on the equator has a centripetal acceleration of \(0.0340 \mathrm{~m} / \mathrm{s}^{2}\), whereas a point at the poles has no centripetal acceleration. (a) Show that, at the equator,the gravitational force on an object (the object's true weight) must exceed the object's apparent weight. (b) What are the apparent weights of a \(75.0-\mathrm{kg}\) person at the equator and at the poles? (Assume Earth is a uniform sphere and take \(g=9.800 \mathrm{~m} / \mathrm{s}^{2}\).)

A \(0.400-\mathrm{kg}\) pendulum bob passes through the lowest part of its path at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). (a) What is the tension in the pendulum cable at this point if the pendulum is \(80.0 \mathrm{~cm}\) long? (b) When the pendulum reaches its highest point, what angle does the cable make with the vertical? (c) What is the tension in the pendulum cable when the pendulum reaches its highest point?

A roller coaster travels in a circular path. (a) Identify the forces on a passenger at the top of the circular loop that cause centripetal acceleration. Show the direction of all forces in a sketch. (b) Identify the forces on the passenger at the bottom of the loop that produce centripetal acceleration. Show these in a sketch. (c) Based on your answers to parts (a) and (b), at what point, top or bottom, should the seat exert the greatest force on the passenger? (d) Assume the speed of the roller coaster is \(4.00 \mathrm{~m} / \mathrm{s}\) at the top of the loop of radius \(8.00 \mathrm{~m}\). Find the force exerted by the seat on a \(70.0-\mathrm{kg}\) passenger at the top of the loop. Then, assume the speed remains the same at the bottom of the loop and find the force exerted by the seat on the passenger at this point. Are your answers consistent with your choice of answers for parts (a) and (b)?

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