All categories

2 years ago

Which components of an atom are found outside of the nucleus

Physics

2 answers:

butalik [34]2 years ago

3 0

Electrons are found outside of the nucleus.

irina [24]2 years ago

3 0

Answer:

electron

Explanation:

I took the test

You can see other question and answer in my article.

Just search in search engine with: Learningandassignments diy4pro

Click on my site and find these related article post:

Atoms Quiz- Grade 8 Unit 2 Leson 2

Hope it helps.

You might be interested in

On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

b) $\Delta s=31868131.8681\ J.K^{-1}$

Explanation:

Given:

amount of heat transfer occurred, $dQ=3.5\times 10^6\ J$

initial temperature of car, $T_i=36.5+273=309.5\ K$

final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

$\Delta s=31868131.8681\ J.K^{-1}$

6 0

3 years ago

(100 pts)<br>Why can't you use distance divide by time to calculate the instantaneous<br>speed?

Fiesta28 [93]

Answer:

Instantaneous speed means speed at any instant

that means Speed is changing with time

You know speed is distance/time

So that means distance is also changing with time

So we take infinitesimal small distance per infinitesimal small time As we assume speed is constant in infinitesimal small time dt

So, we take speed = ds/dt

ds = infinitesimal small distance

dt = infinitesimal small time

As its ratio is equal to speed at any instant

Note : We are taking infinitesimal small distance

But :) we are taking infinitesimal small time also

As you know if denominator is small fraction is large So fraction always give large value

So it's not O ( this makes confuse to most of students)

So, thanks

Good question

Keep thinking like this :)

4 0

3 years ago

Read 2 more answers

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the

scoundrel [369]

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = $\frac{kQ_{1}Q_{2} }{r^{2} }$

F= $\frac{K(Q/2)(3Q/4)}{r^{2} }$

how $\frac{KQ^{2} }{r^{2} }$ = 0.42

Then

F = $\frac{0.42*3}{8}$

F = 0.1575 N

3 0

3 years ago

A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energ

lesantik [10]

Answer:

- 2.7 x 10^-6 J

Explanation:

q1 = 1 nC at x = 0 cm

q2 = - 1 nC at x = 1 cm

q3 = 4 nC at x = 2 cm

The formula for the potential energy between the two charges is given by

$U=\frac{Kq_{1}q_{2}}{r}$

where r be the distance between the two charges

By use of superposition principle, the total energy of the system is given by

$U = U_{1,2}+U_{2,3}+U_{3,1}$

$U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}$

$U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}$

U = - 2.7 x 10^-6 J

3 0

2 years ago

The mass of earth 6*10^24kg and radius is 400kg Now find the value of acceleration due to gravity when a object is 3600km from t

vlada-n [284]

I don’t worry wewwwww it is a good time to get it done lol lol i don’t worry about it lol lol i lol

5 0

2 years ago