/*! 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

91Ó°ÊÓ

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
The centrifuge should operate at approximately \(10.623\) revolutions per second.

Step by step solution

01

Calculate the Acceleration

Using the equation of motion F = ma, where F is the force, m is the mass, and a is the acceleration, we can find the acceleration of the red blood cell. We have F= \(4.0 \times 10^{-11}\) N and m= \(3.0 \times 10^{-16}\) kg. By substituting these values into the equation, we get a = F/m = \(4.0 \times 10^{-11} N / 3.0 \times 10^{-16} kg = 1.33 \times 10^5 m/s^2\).
02

Find the Angular Velocity

The formula for centripetal acceleration in terms of angular velocity is a = ω^2r. Here a is the acceleration found from Step 1, r is the radius of the centrifuge, and ω is the angular velocity. We know the radius r = 0.15 m (converted from cm to m), and we're solving for ω. Rearranging the formula gives ω = sqrt(a/r) = sqrt(\(1.33 \times 10^5 m/s^2) / 0.15 m = 66.746 rad/s.\)
03

Convert Rad/s to Revolutions/s

To convert from rad/s to revolutions/s, we divide the value of ω by \(2π\), because there are \(2π\) radians in a revolution. This gives revolutions per second equals to ω / \(2π\)= 66.746 rad/s / \(2π\) = 10.623 rev/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Centripetal Acceleration
The concept of centripetal acceleration is crucial when studying objects moving in a circular path. It is the acceleration directed towards the center of the circle that keeps an object moving along the circular path. Even if the speed is constant, the velocity is not, because velocity is a vector and has both magnitude and direction. In this exercise, the centripetal acceleration is critical to determining the speed at which a centrifuge must spin to apply the necessary force to separate blood cells.

Using the formula for centripetal force, which is expressed mathematically as \( F = m \times a \), where \( F \) represents the force, \( m \) stands for the mass of the object, and \( a \) symbolizes the centripetal acceleration. From the given values, we ascertain the centripetal acceleration by rearranging the formula as \( a = F/m \). The resulting acceleration indicates how sharply the blood cell is changing its velocity, aiming towards the center of the centrifuge's path.
Angular Velocity
Angular velocity relates to objects moving in a circular path and measures how fast an object rotates or revolves relative to another point, usually the center of a circle. It is quantified as an angle rotated per unit time and is typically expressed in radians per second (rad/s). In our context, angular velocity is related to centripetal acceleration through the formula \( a = \omega^2r \), where \( \omega \) is the angular velocity, and \( r \) is the radius of the circular path.

In the given exercise, finding the angular velocity (\( \omega \)) allows us to understand how fast the centrifuge spins. After computing the centripetal acceleration, we can rearrange the formula to \( \omega = \sqrt{a/r} \), giving us the angular velocity in radians per second based on the calculated acceleration and the radius of the centrifuge. This value is a significant step in determining the operational speed for the centrifuge that influences the separation process of the blood cells.
Revolutions Per Second
The term 'revolutions per second' is a measure of rotational speed commonly used to describe how many complete turns an object makes in one second. It is directly related to the angular velocity but expressed in units of cycles or revolutions, which can be more intuitive. To convert from radians per second to revolutions per second, we divide the angular velocity by \(2\pi\), as there are \(2\pi\) radians in one full revolution.

From the solution provided, we understand how this conversion is applied, \( \omega / 2\pi = 10.623 \text{ rev/s} \). This final value is crucial in a practical setting when we control the centrifuge; it provides an easy-to-understand number that the equipment can be set to, ensuring the correct force is applied to efficiently separate the blood cells from the plasma in the sample. This conversion facilitates the usage of centrifuges, as their operational settings often require input in revolutions per minute (RPM) or per second.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

ecp A minimum-energy orbit to an outer planet consists of putting a spacecraft on an elliptical trajectory with the departure planet corresponding to the perihelion of the ellipse, or closest point to the Sun, and the arrival planet corresponding to the aphelion of the ellipse, or farthest point from the Sun. (a) Use Kepler's third law to calculate how long it would take to go from Earth to Mars on such an orbit. (Answer in years.) (b) Can such an orbit be undertaken at any time? Explain.

A certain light truck can go around a flat curve having a radius of \(150 \mathrm{~m}\) with a maximum speed of \(32.0 \mathrm{~m} / \mathrm{s}\). With what maximum speed can it go around a curve having a radius of \(75.0 \mathrm{~m}\) ?

The diameters of the main rotor and tail rotor of a singleengine helicopter are \(7.60 \mathrm{~m}\) and \(1.02 \mathrm{~m}\), respectively. The respective rotational speeds are 450 rev/min and 4138 rev/min. Calculate the speeds of the tips of both rotors. Compare these speeds with the speed of sound, \(343 \mathrm{~m} / \mathrm{s}\)

(a) What is the tangential acceleration of a bug on the rim of a 10 -in.-diameter disk if the disk moves from rest to an angular speed of 78 rev/min in \(3.0\) s? (b) When the disk is at its final speed, what is the tangential velocity of the bug? (c) One second after the bug starts from rest, what are its tangential acceleration, centripetal acceleration, and total acceleration?

In Robert Heinlein's The Moon Is a Harsh Mistress, the colonial inhabitants of the Moon threaten to launch rocks down onto Earth if they are not given independence (or at least representation). Assuming a gun could launch a rock of mass \(m\) at twice the lunar escape speed, calculate the speed of the rock as it enters Earth's atmosphere.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.