/*! 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} Q29P Suppose a gangster sprays Superm... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Q29P (page 249)

Suppose a gangster sprays Superman鈥檚 chest with3 gbullets at the rate of 100 bullets/min, and the speed of each bullet is 500 m/s. Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman鈥檚 chest?

Short Answer

Expert verified

The magnitude of the average force on Superman鈥檚 chest is Favg=5.0N

Step by step solution

01

Understanding the given information

The mass of each bullet,m=3g1kg1000g=3脳10-3kg

The speed of a bullet is,v=500m/s

The rate of bombarding bullets is, 100 bullets/min .

02

Concept and formula used in the given question

We can use the equation of impulse-momentum theorem to calculate the impulse and then we can use this value of impulse in the equation of average force to calculate its value.

J=鈭�P鈭�P=m脳v2-v1J=Favg脳t

03

Calculation for the magnitude of the average force on Superman’s chest

Suppose, the time of bombardingthebullet is 1 min. So, the total number of bullets shot at is 100.

The equation of the impulse-momentum theorem is,

J=鈭�PJ=100脳m脳v2-v1J=100脳3脳10-3kg脳500m/s--500m/sJ=300N路s

Now,

Favg=J鈭�t=300N路s60s=5.0N

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

In Fig. 9-21, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The alpha particle is scattered at angle 胃1=64.0掳and the oxygen nucleus recoils with speed localid="1661254577203" 1.20脳105m/sand at angle 胃1=51.0掳.In atomic mass units, the mass of the alpha particle is 4.00 uand the mass of the oxygen nucleus is 16.0 u. What are the (a) final and (b) initial speeds of the alpha particle?

A collision occurs between a 2.00 kgparticle travelling with velocity v1=(-4.00ms)i铃�+(-5.00ms)j铃�and a 4.00 kgparticle travelling with velocity v2=(6.00ms)i铃�+(-2.00ms)j铃�. The collision connects the two particles. What then is their velocity in (a) unit-vector notation and as a (b) Magnitude and (c) Angle?

Basilisk lizards can run across the top of a water surface (Figure 9-52). With each step, a lizard first slaps its foot against the water and then pushes it down into the water rapidly enough to form an air cavity around the top of the foot. To avoid having to pull the foot back up against water drag in order to complete the step, the lizard withdraws the foot before water can flow into the air cavity. If the lizard is not to sink, the average upward impulse on the lizard during this full action of slap, downward push, and withdrawal must match the downward impulse due to the gravitational force. Suppose the mass of a basilisk lizard is 9.00 g, the mass of each foot is 3.00 g , the speed of a foot as it slaps the water is 1.50 m/s , and the time for a single step is 0.600 s .(a) What is the magnitude of the impulse on the lizard during the slap? (Assume this impulse is directly upward.) (b) During the 0.600 sduration of a step, what is the downward impulse on the lizard due to the gravitational force? (c) Which action, the slap or the push, provides the primary support for the lizard, or are they approximately equal in their support?

A force in the negative direction of an x-axis is applied for 27 msto a 0.40 kgball initially moving at14 m/sin the positive direction of the axis. The force varies in magnitude, and the impulse has magnitude32.4 N.s. What are the balls (a) Speed and (b) Direction of travel just after the force is applied? What are (c) The average magnitude of the force and (d) The direction of the impulse on the ball?

A spacecraft is separated into two parts by detonating the explosive bolts that hold them together. The masses of the parts are 1200 kgand 1800 kg; the magnitude of the impulse on each part from the bolts is 300 N. With what relative speed do the two parts separate because of the detonation?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Waves Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Engineering Physics

Read Explanation

Energy Physics

Read Explanation

Astrophysics

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.