Physics : asked on csulli2006
 08.12.2022

A class is performing an experiment to keep an ice cube solid as long as possible.

. 8

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Physics
Step-by-step answer
P Answered by PhD

B. Maria wrapped the ice cube tightly in cotton and aluminum foil.

Explanation:

According to this question, a class is performing an experiment to keep an ice cube solid for as long as possible. However, in the experimental procedure that will be used to actualize this aim, certain protocols must be duly observed to avoid error in the experiment.

Among the listed options containing actions taken by students, wrapping the ice cube tightly in cotton and aluminum foil is a likely source of error to the experiment. This is because aluminum, owing to its metallic properties, is a good conductor of heat. Hence, Aluminum foil will conduct heat and cause the ice cube to melt faster than normal.

This will jeopardize the aim of the experiment, which is to keep an ice cube solid for as long as possible.

Physics
Step-by-step answer
P Answered by PhD

B. Maria wrapped the ice cube tightly in cotton and aluminum foil.

Explanation:

According to this question, a class is performing an experiment to keep an ice cube solid for as long as possible. However, in the experimental procedure that will be used to actualize this aim, certain protocols must be duly observed to avoid error in the experiment.

Among the listed options containing actions taken by students, wrapping the ice cube tightly in cotton and aluminum foil is a likely source of error to the experiment. This is because aluminum, owing to its metallic properties, is a good conductor of heat. Hence, Aluminum foil will conduct heat and cause the ice cube to melt faster than normal.

This will jeopardize the aim of the experiment, which is to keep an ice cube solid for as long as possible.

English
Step-by-step answer
P Answered by PhD

A) Your years in college are some of the best years of your life

Explanation:

So, this article deals with some advice regarding college years, what to do and what not to do.

Basically, college presents a fresh start, chance for making new friends, experiences, seeing new places, college years serve to prepare one for an adult life.

However, the author doesn't claim it well be all easy and fun; taking classes, studying for exams, having a part time job can really be exhausting.

Despite of that, the author's primary opinion can best be seen in the conclusion: "For once, the cliché you’ve heard your entire high school career is true—college years are some of the best years of your life"

English
Step-by-step answer
P Answered by PhD

A) Your years in college are some of the best years of your life

Explanation:

So, this article deals with some advice regarding college years, what to do and what not to do.

Basically, college presents a fresh start, chance for making new friends, experiences, seeing new places, college years serve to prepare one for an adult life.

However, the author doesn't claim it well be all easy and fun; taking classes, studying for exams, having a part time job can really be exhausting.

Despite of that, the author's primary opinion can best be seen in the conclusion: "For once, the cliché you’ve heard your entire high school career is true—college years are some of the best years of your life"

Physics
Step-by-step answer
P Answered by Master

Answer:

see below.

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

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