/*! 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} Q112P The speed of a bullet is measure... [FREE SOLUTION] | 91影视

91影视

The speed of a bullet is measured to be 640m/s as the bullet emerges from a barrel of length 1.20m. Assuming constant acceleration, find the time that the bullet spends in the barrel after it is fired.

Short Answer

Expert verified

The time spent by the bullet in barrel after it is fired is3.76ms.

Step by step solution

01

Given data

The final speed of the bullet,v=640m/s

The length of the barrel, x=1.20m

02

Understanding the kinematic equations

Kinematic equations describe the motion of an object with constant acceleration. These equations can be used to determine the acceleration, velocity or distance.

The expression for the kinematic equations of motion are given as follows:

v=v0+at 鈥 (i)

v2=v02+2ax 鈥 (ii)

Here, v0 is the initial velocity, vis the final velocity, tis the time, ais the acceleration andxis the displacement.

03

Determination of the acceleration of bullet.

Since the bullet starts from rest, the initial speed is zero, that is,v0=0m/s .

Using equation (ii), the acceleration can be calculated as follows:

a=v2-v022x=(640m/s)2-(0m/s)221.20m=1.7105m/s2

04

Determination of the time spent by the bullet in barrel

Using equation (i), the time spent by the bullet is calculate as follows:

t=v-v0a=640m/s-0m/s1.710-3s=3.76m/s

Therefore, the time spent by the bullet in the barrel after it is fired is3.76ms .

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

A rock is thrown vertically upward from ground level at time t=0 . Atrole="math" localid="1656149217888" t=1.5s ,it passes the top of a tall tower, and 1.0 s later, it reaches its maximum height. What is the height of the tower?

Figure 2-35 shows the speed v versus height y of a ball tossed directly upward, along a y axis. Distance d is0.40 m. The speed at heightyAisvA.The speed at heightyBis13VA. What is speedVA?

Inraces, runner 1 on track 1 (with time 2min,27.95sec) appears to be faster than runner 2 on track 2 (2min,28.15sec). However, length L2of track 2 might be slightly greater than length L1of track 1. How large L2-L1can be for us still to conclude that runner 1 is faster?

The wings on a stonefly do not flap, and thus the insect cannot fly. However, when the insect is on a water surface, it can sail across the surface by lifting its wings into a breeze. Suppose that you time stoneflies as they move at constant speed along a straight path of a certain length. On average, the trips each take7.1swith the wings set as sails and 25.0swith the wings tucked in. (a) What is the ratio of the sailing speed vsto the non-sailing speed localid="1654754209433" vns? (b) In terms oflocalid="1654754226410" vs, what is the difference in the times the insects take to travel the first2.0malong the path with and without sailing?

A motorcyclist who is moving along an x axis directed toward the east has an acceleration given by a=(6.11.2t)m/s2for0t6.0s. Att=0, the velocity and position of the cyclist are2.7m/s and7.3m. (a) What is the maximum speed achieved by the cyclist? (b) What total distance does the cyclist travel betweent=0and6.0s?

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.