/*! 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} Q. 10 FIGURE EX2.7 showed the velocity... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q. 10 (page 59)

FIGURE EX2.7 showed the velocity graph of blood in the aorta. What is the blood鈥檚 acceleration during each phase of the motion, speeding up and slowing down?

Short Answer

Expert verified

The acceleration of blood during the speeding up phase is 16ms-2.

The acceleration of blood during the speeding down phase is-5.3ms-2.

Step by step solution

01

Given information

The velocity-time graph of blood in the aorta is shown below

02

Find the acceleration of blood in the aorta during the speeding up phase.

Average acceleration is given by the slope of the velocity-versus-time graph.

For the first phase, consider two points t1,v1=0.1,0and t2,v2=0.15,0.8.

The acceleration of the blood is

a=v2-v1t2-t1=0.8-00.15-0.1=16ms2

Therefore, the acceleration of blood during the speeding up phase is16ms2.

03

Find the acceleration of blood in the aorta during the speeding down phase.

For the second phase, consider two points t2,v2=0.15,0.8and t3,v3=0.3,0.

The acceleration of the blood is

a=v3-v2t3-t2=0-0.80.3-0.15=-5.3ms2

Therefore, the acceleration of blood during the speeding down phase is-5.3ms2.

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 91影视!

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

You are playing miniature golf at the golf course shown in

FIGURE P2.51. Due to the fake plastic grass, the ball decelerates at 1.0 m/s2 when rolling horizontally and at 6.0 m/s2 on the slope. What is the slowest speed with which the ball can leave your golf club if you wish to make a hole in one?

a. How many days will it take a spaceship to accelerate to the speed of light ( 13.0 * 108 m/s) with the acceleration g? b. How far will it travel during this interval? c. What fraction of a light-year is your answer to part b? A light-year is the distance light travels in one year

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at a height h directly above the first ball is dropped from rest.
a. At what height above the ground do the balls collide? Your answer will be an algebraic expression in terms of h, v0, and g.
b. What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
c. For what value of h does the collision occur at the instant when the first ball is at its highest point?

FIGURE P2.45 shows a set of kinematic graphs for a ball rolling on a track. All segments of the track are straight lines, but some may be tilted. Draw a picture of the track and also indicate the ball鈥檚 initial condition.

A sprinter can accelerate with constant acceleration for 4.0 s before reaching top speed. He can run the 100 meter dash in 10.0 s. What is his speed as he crosses the finish line?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Electromagnetism

Read Explanation

Thermodynamics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Oscillations

Read Explanation

Wave Optics

Read Explanation
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.