EBK LOOSE-LEAF VERSION OF UNIVERSE
EBK LOOSE-LEAF VERSION OF UNIVERSE
11th Edition
ISBN: 9781319227975
Author: KAUFMANN
Publisher: VST
bartleby

Videos

Question
Book Icon
Chapter 15, Problem 47Q

(a)

To determine

The mass of the nucleus of a comet with side length 10km and density 1000kg/m3.

(a)

Expert Solution
Check Mark

Answer to Problem 47Q

Solution:

1015kg.

Explanation of Solution

Given data:

The side length of the comet’s nucleus is 10km and the density of the comet is 1000kg/m3.

Formula used:

Write the expression for density.

ρ=mV

Here, m and V are, respectively, the mass and density of the comet’s nucleus.

Write the expression for the volume for a cube.

V=a3

Here, a is the side length of the cube.

Explanation:

Recall the expression for the mass density of the comet’s nucleus.

ρ=mV

Rearrange the above expression for m.

m=ρV

Since the expression for the volume of the comet’s nucleus is V=a3, the above formula becomes,

m=ρa3

Substitute 1000kg/m3 for ρ and 10km for a in the above expression.

m=1000kg/m3(10 km(103 m1 km))3=1000×1012 kg=1015kg

Conclusion:

Hence, the mass of the nucleus of the comet is 1015kg.

(b)

To determine

The average density of the tail of the comet which has a length of 108km and a width of 106km, when 1% of the mass of the nucleus of the comet evaporates to form the tail. Also, take the air density as 1.2 kg/m3.

(b)

Expert Solution
Check Mark

Answer to Problem 47Q

Solution:

1011kg/m3

Explanation of Solution

Given data:

The length of the tail is 108km, width of the tail is 106km, the mass of the tail is 1% of the mass of the comet’s nucleus and air density is 1.2 kg/m3.

Formula used:

Write the expression for density.

ρ=mV

Here, m and V are, respectively, the mass and density of the comet’s nucleus.

Explanation:

The mass of the comet’s nucleus was found in the previous part to be 1015kg and it is given that mass of the tail is 1% of the mass of the nucleus.

mtail=1%ofm=1(1100×1015 kg)=1013kg

Now, volume of the tail can be calculated as:

Vtail=l×b×h

Here, landb are, respectively, the length and width of the tail and h is the side length of the comet’s nucleus.

Substitute 108km for l, 106km for b and 10km for h.

Vtail=108 km(106 km)(10 km)=1015km3((103 m1 km)3)=1024m3

Thus, recall the expression for the density of the tail.

ρtail=mtailVtail

Here, mtail, ρtail and Vtail are, respectively, the mass, volume and density of the comet’s tail.

Substitute 1013kg for mtail and 1024m3 for Vtail.

ρtail=1013 km 1024m3=1011kg/m3

Conclusion:

Hence, the average density of the tail of the comet is 1011kg/m3.

(c)

To determine

Whether the passing of the comet’s tail through Earth leads to harmful effects on the health of human beings.

(c)

Expert Solution
Check Mark

Answer to Problem 47Q

Solution:

The tail of the comet does not affect the health of human beings on Earth.

Explanation of Solution

Introduction:

The air we breathe has a density of about 1.2kg/m3. Any gas mixed with air that does not change the density of air much will not harm the human body.

Explanation:

The density of the tail of the comet is 1011kg/m3, which is very low compared to the density of the air we breathe. When Earth passes through the tail, the air in Earth’s atmosphere also mixes with the tail, but the mixed air is not harmful for human beings to breathe.

Conclusion:

Hence, the passing of the Earth through the comet’s tail is not harmful for human life.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is the same as that between the ladder and the wall. If this coefficient of static friction is μs : 0.535, determine the smallest angle the ladder can make with the floor without slipping. ° = A 14.0 m uniform ladder weighing 480 N rests against a frictionless wall. The ladder makes a 55.0°-angle with the horizontal. (a) Find the horizontal and vertical forces (in N) the ground exerts on the base of the ladder when an 850-N firefighter has climbed 4.10 m along the ladder from the bottom. horizontal force magnitude 342. N direction towards the wall ✓ vertical force 1330 N up magnitude direction (b) If the ladder is just on the verge of slipping when the firefighter is 9.10 m from the bottom, what is the coefficient of static friction between ladder and ground? 0.26 × You appear to be using 4.10 m from part (a) for the position of the…
Your neighbor designs automobiles for a living. You are fascinated with her work. She is designing a new automobile and needs to determine how strong the front suspension should be. She knows of your fascination with her work and your expertise in physics, so she asks you to determine how large the normal force on the front wheels of her design automobile could become under a hard stop, ma when the wheels are locked and the automobile is skidding on the road. She gives you the following information. The mass of the automobile is m₂ = 1.10 × 103 kg and it can carry five passengers of average mass m = 80.0 kg. The front and rear wheels are separated by d = 4.45 m. The center of mass of the car carrying five passengers is dCM = 2.25 m behind the front wheels and hcm = 0.630 m above the roadway. A typical coefficient of kinetic friction between tires and roadway is μk 0.840. (Caution: The braking automobile is not in an inertial reference frame. Enter the magnitude of the force in N.)…
John is pushing his daughter Rachel in a wheelbarrow when it is stopped by a brick 8.00 cm high (see the figure below). The handles make an angle of 0 = 17.5° with the ground. Due to the weight of Rachel and the wheelbarrow, a downward force of 403 N is exerted at the center of the wheel, which has a radius of 16.0 cm. Assume the brick remains fixed and does not slide along the ground. Also assume the force applied by John is directed exactly toward the center of the wheel. (Choose the positive x-axis to be pointing to the right.) (a) What force (in N) must John apply along the handles to just start the wheel over the brick? (No Response) N (b) What is the force (magnitude in kN and direction in degrees clockwise from the -x-axis) that the brick exerts on the wheel just as the wheel begins to lift over the brick? magnitude (No Response) KN direction (No Response) ° clockwise from the -x-axis
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Astronomy
Physics
ISBN:9781938168284
Author:Andrew Fraknoi; David Morrison; Sidney C. Wolff
Publisher:OpenStax
Text book image
Horizons: Exploring the Universe (MindTap Course ...
Physics
ISBN:9781305960961
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
The Solar System
Physics
ISBN:9781305804562
Author:Seeds
Publisher:Cengage
Text book image
Foundations of Astronomy (MindTap Course List)
Physics
ISBN:9781337399920
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Text book image
Stars and Galaxies (MindTap Course List)
Physics
ISBN:9781337399944
Author:Michael A. Seeds
Publisher:Cengage Learning
Kepler's Three Laws Explained; Author: PhysicsHigh;https://www.youtube.com/watch?v=kyR6EO_RMKE;License: Standard YouTube License, CC-BY