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Alenkasestr [34]
2 years ago
10

1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True B. False

Physics
2 answers:
Studentka2010 [4]2 years ago
4 0
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True

2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is? 
B.6

3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.
ohaa [14]2 years ago
3 0

Answer:

1.True

2.6

3.All of the above

Explanation:

1.In a single atom,  no more than 2 electrons can occupy a single orbital.

By pauli exclusion principle, an orbital can contain a maximum of only 2 electrons, the two electrons must have opposite spins.

Therefore, it is true.

2.We know that maximum number of electrons in p sublevel is 6.

Hence, the maximum number of electrons allowed in a p sublevel of the 3rd principle level is 6.

3.A neutral atom has a ground state electronic configuration of 1s^2 2s^2.

Therefore, Total number of electrons=4

Atomic number=4

Number of protons=Atomic number=4

The atom has  total number of orbital is 2.

The element has no unpaired electrons.

Hence, all of the above is correct option.

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A ball with an initial velocity of 2 m/s rolls for a period of 3 seconds. If the ball is uniformly accelerating at a rate of 3 m
ikadub [295]

Answer: 11 m/s

vinitial=2 m/s

time=3 s

acceleration = 3 m/s^2

vfinal = ?

The key here is that it is a constant acceleration, so we can use the constant acceleration equations. The easiest one to use would be:

vfinal=vinitial + a*t

We need vfinal, so algebraically we are ready to put in numbers into the equation:

vfinal=vinitial + a*t = 2 m/s + (3 m/s^2)*(3 s ) = 11 m/s is the final velocity

7 0
2 years ago
A boy does 465 J of work pulling an empty wagon along level ground with a force of 111 N [31o below the horizontal]. A frictiona
vovikov84 [41]
Can you list the answer's
6 0
2 years ago
A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later
mart [117]

Explanation:

It is given that,

The horizontal speed of a cliff diver, v_x=3.6\ m/s

It reaches the water below 2.00 s later, t = 2 s

Let d_x is the distance where the diver hit the water. It can be calculated as follows :

d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m

Let d_y is the height of the cliff. It can be calculated using second equation of motion as follows :

d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

8 0
1 year ago
A) The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion
suter [353]

A) The mass of the continent is 2.5\cdot 10^{21} kg

B) The kinetic energy is 2016 J

C) The speed of the jogger should be 7.1 m/s

Explanation:

A)

The mass of the continent can be calculated as

m = \rho V

where

\rho = 2800 kg/m^3 is its density

V is its volume

We have to calculate its volume. We know that the continent is represented as a slab of side 5900 km (so its surface is 5900 x 5900, assuming it is a square) and depth of 26 km, so its volume is:

V=(5900 km)^2 (26 km)=9.05\cdot 10^8 km^3 =9.05 \cdot 10^8 \cdot (10^9 m^3/k^3)=9.05\cdot 10^7 m^3

So, the mass of the continent is

m=\rho V = (2800)(9.05\cdot 10^{17})=2.5\cdot 10^{21} kg

B)

The kinetic energy of a body is given by

K=\frac{1}{2}mv^2

where

m is the mass of the body

v is its speed

For the continent, we have:

m=2.5\cdot 10^{21} kg is the mass

v=4 cm/year is the speed

We have to convert the speed into SI units. we have:

1 cm = 0.01 m

1 year = (365)(24)(60)(60) s = 3.15\cdot 10^7 s

So, the speed is

v=4 cm/year = 0.04 m/year \cdot \frac{1}{3.15\cdot 10^7}=1.27\cdot 10^{-9} m/s

Therefore, the kinetic energy is

K=\frac{1}{2}(2.5\cdot 10^{21} kg)(1.27\cdot 10^{-9} m/s)^2=2016 J

C)

Again, the kinetic energy of an object is

K=\frac{1}{2}mv^2

For the jogger in this problem, his mass is

m = 80 kg

And we want its kinetic energy to be equal to that of the continent, so

K = 2016 J

Re-arranging the equation for v, we find what speed the jogger needs to have this kinetic energy:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2016)}{80}}=7.1 m/s

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

8 0
2 years ago
If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be?
kicyunya [14]

For this case we have that by definition:

F = ma

Where,

  • <em>m: mass of the object </em>
  • <em>a: acceleration of the object </em>
  • <em>F: summation of forces </em>

We have then:

F = 10 + 5\\F = 15 N

Then, by clearing the acceleration we have:

a = \frac {F} {m}

Substituting values we have:

a = \frac {15} {5}\\a = 3 \frac {m} {s ^ 2}

Answer:

The acceleration of the box is equal to:

a = 3 \frac {m} {s ^ 2}

6 0
2 years ago
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