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2 years ago

The maximum number of orbits in an atom is _____

Physics

1 answer:

svp [43]2 years ago

8 0

Correct answer is:

<h2>The maximum number of orbits in an atom is <u>Seven.</u></h2><h3>Explanation:</h3>

Every energy level has a limited one orbital including two electrons. The orbits are settled in the sub-levels and there can be further than 1 sub-level as the number of energy levels rises. On energy level 1, there is 1 sub-level and 1 orbital. Energy level 2 can possess 2 sub-levels and 2 orbitals. These remain to develop as you progress from the nucleus of the atom, closing up with an infinite potential number of levels and orbits.

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Why does the band of stability curve upward at high atomic numbers? excess neutrons are required due to the repulsion between th

ella [17]

The band of stability curves upward at high atomic numbers due to the fact that excess of neutrons are required due to the repulsion between protons.

Atomic number is the number of protons. As the number of protons (atomic number) increase, the electrical repulsion force, due to the same sign of the protons inside the nucleus, becomes more dominant compared to the nuclear force attractions, then the nucleus needs more neutrons to gain stability.The presence of more neutrons decrease the density of protons which decreases the repulsion among them.

3 0

3 years ago

The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps an

Marrrta [24]

Answer: $k=55.590 KN/m$

Explanation:

Given

mass of person $\left ( m\right )$ =68 kg

car dips about 1.2 cm

We know

F=kx

Where k=combined spring constant

mg=kx

$k=\frac{mg}{x}$

$k=\frac{68\times 9.81}{1.2\time 10^{-2}}$

$k=55.590 KN/m$

7 0

3 years ago

An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi

shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = $a^{2}$

Let the height of the bin be 'h'

Therefore the total area, $A_{t} = 4ah$

The cost is:

C = 2sh

Volume of the box, V = $a^{2}h = 4^{2}\times 2 = 128$ (1)

Total cost, $C_{t} = 2a^{2} + 2ah$ (2)

From eqn (1):

$h = \frac{128}{a^{2}}$

Using the above value in eqn (1):

$C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}$

$C(a) = 2a^{2} + \frac{256}{a}$

Differentiating the above eqn w.r.t 'a':

$C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}$

For the required solution equating the above eqn to zero:

$\frac{4a^{3} - 256}{a^{2}} = 0$

a = 4

Also

$h = \frac{128}{4^{2}} = 8$

The path in order to minimize the cost must be a rectangle.

8 0

3 years ago

Which of the following statements are true? Select all correct responses. Choose all that apply. Choose all that apply. The tota

Ne4ueva [31]

<h3><u>Answer</u>;</h3>

-The total momentum of an isolated system is constant.

-The total momentum of any number of particles is equal to the vector sum of the momenta of the individual particles.

-The vector sum of forces acting on a particle equals the rate of change of momentum of the particle with respect to time.

<h3><u>Explanation</u>;</h3>

Momentum is a vector quantity, and therefore we need to use vector addition when summing together the momenta of the multiple bodies which make up a system.
The vector sum of forces acting on a particle is equivalent to the rate of change of momentum of the particle with respect to time. This is according to the Newton's second Law of motion. In mathematical terms, ֿF = d ֿp/dt, that is F= ma.
According to the Law of conservation of Momentum, or a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

3 0

3 years ago

light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

lapo4ka [179]

Answer:

The distance traveled in 1 year is: $3.143*10^{16}ft$