A light-year is about?

An atom with the expected number of neutrons, protons, and electrons is called a(n)

ryzh [129]

Answer:

Stable atom

Explanation:

A stable atom is one that has a balanced nuclear inter-particle force reaction as such the binding energy of a stable atom is sufficient to permanently keep the nucleus as one unit. Examples of a stable atom are the atoms of monoisotopic elements such as fluorine, sodium, iodine, gold, aluminium, and cobalt.

In a stable atom the expected number of proton, neutron, and electron are present while in an unstable atom or radioactive atom, there are more than the expected number of neutrons or protons, such that the internal energy of the nucleus is excessive and more than the binding energy, which can lead to radioactive decay.

6 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

Why do the four outer planets have lower density than the four inner planets?

Marrrta [24]

Because the four outer planets are comprised of mostly gases give me a thanks or brainiest answer if this helps!

4 0

3 years ago

A stable, binary ionic compound exists where the cation (A) has a charge twice in magnitude than that of the anion (X). What wou

svetlana [45]

The cation is positively charged and has a charge twice that of the anion, for example +2. The anion is negatively charged and in our example where the cation has a +2 charge, it must have a -1 charge. In order for the charges to equal zero, there must be two anions: -1 x 2 = -2 So the answer is D. AX2

8 0

3 years ago

Read 2 more answers

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback ca

love history [14]

Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. $s_{o}$ = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.