/*! 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} Q92PP The velocity of blood in the aor... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The velocity of blood in the aorta can be measured directly with ultrasound techniques. A typical graph of blood velocity versus time during a single heartbeat is shown in Fig. P2.92. Which statement is the best interpretation of this graph? (a) The blood flow changes direction at about 0.25 s; (b) the speed of the blood flow begins to decrease at about 0.10 s; (c) the acceleration of the blood is greatest in magnitude at about 0.25 s; (d) the acceleration of the blood is greatest in magnitude at about 0.10 s.

Short Answer

Expert verified

The acceleration of the blood is greater in magnitude at about 0.10s.

Step by step solution

01

Identification of given data

The time intervals at which blood velocity is shown is,0.10s,0.25s,0.50s,0.75s,1.00s.

The given blood velocity is,0.2m/s,0.4m/s,0.6m/s,0.8m/s,1.0m/s

02

Concept of instantaneous acceleration

The ratio of velocity changes during a certain time period to zero.

The instantaneous acceleration is given by,

a=ddtdxdt=d2xdt2...i

03

Evaluate the velocity-time graph of blood in aorta

The direction of blood flow does not alter.The rate of blood flow decreases.

After 0.2s, the slope flattens off and blood flow speeds up the most.

The slope of graph is steeper about0.10s ,where acceleration of blood is greatest.

Thus, the acceleration of the blood is greatest about 0.10s.

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!

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

A Simple Reaction-Time Test.A meter stick is held vertically above your hand, with the lower end between your thumb and first finger. When you see the meter stick released, you grab it with those two fingers. You can calculate your reaction time from the distance the meter stick falls, read directly from the point where your fingers grabbed it. (a) Derive a relationship for your reaction time in terms of this measured distance, d. (b) If the measured distance is 17.6 cm, what is your reaction time?

A circular racetrack has a radius of 500 m. What is the displacement of a bicyclist when she travels around the track from the north side to the south side? When she makes one complete circle around the track? Explain.

The following conversions occur frequently in physics and are very useful. (a) Use 1 mi = 5280 ft and 1 h = 3600 s to convert 60 mph to units of ft/s. (b) The acceleration of a freely falling object is 32 ft/s2. Use 1 ft = 30.48 cm to express this acceleration in units of m/s2. (c) The density of water is 1.0 g/cm3. Convert this density to units of kg/m3.

A physics professor leaves her house and walks along the sidewalk toward campus. After 5 min, it starts to rain, and she returns home. Her distance from her house as a function of time is shown in Fig. E2.10 At which of the labeled points is her velocity (a) zero? (b) constant and positive? (c) constant and negative? (d) increasing in magnitude? (e) decreasing in magnitude?

If C→=A→+B→ ,what must be true about the directions and magnitudes of A→ and B→ifC=A+B ? What must be true about the directions and magnitudes of role="math" localid="1647990633202" A→andB→ ifC=0 ?

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.