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Svetllana [295]
3 years ago
13

Argon, neon, and xenon are examples of _________. A) halogens B) metalloids C) noble gases D) alkali metals

Physics
2 answers:
Ronch [10]3 years ago
5 0

Answer:

Argon, neon and xenon are examples of noble gases.

C is correct option.

Explanation:

According to electronic configuration

Argon:

The atomic number of argon is 18.

The electronic configuration of argon

Ar= 2,8,8

Ar = 1s^2,2s^22p^6,3s^23p^6

Neon:

The atomic number of argon is 10.

The electronic configuration of argon

Ar= 2,8

Ne = 1s^2,2s^22p^6

Xenon:

The atomic number of argon is 54.

The electronic configuration of argon

Ar= 2,8,18,18,8

Ne = 1s^2,2s^22p^6,3s^23p^6 3d^{10},4s^24p^6 4d^{10},5s^2 5p^6

Argon, neon and xenon are examples of noble gases because the last shell is completely filled and they does not participate in any reaction so, these atoms are called noble gasses.

Hence, Argon, neon and xenon are examples of noble gases.

Leona [35]3 years ago
4 0
C) Noble gases! They are all in the last column of the periodic table!
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1. An object is projected upward with a velocity of 125 m/s.
grigory [225]

Answer:

A) s = 796.38 m

B) t = 12.742 s

C) T = 25.484 s

Explanation:

A) First of all let's find the time it takes to get to maximum height using Newton's first equation of motion.

v = u + gt

u = 125 m/s

v = 0 m/s

g = 9.81 m/s²

Thus;

0 = 125 - 9.81(t)

g is negative because motion is against gravity. Thus;

9.81t = 125

t = 125/9.81

t = 12.742 s

Max height will be gotten from Newton's 2nd equation of motion;

s = ut + ½gt²

s = (125 × 12.742) + (½ × -9.81 × 12.742²)

s = 1592.75 - 796.37

s = 796.38 m

B) time to reach maximum height is;

t = u/g

t = 125/9.81

t = 12.742 s

C) Total time elapsed is;

T = 2u/g

T = 2 × 125/9.81

T = 25.484 s

4 0
3 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
Nina [5.8K]

Answer:

1 × 10⁶ N/C

Explanation:

The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C

4 0
3 years ago
At the end of cylindrical rod of length l = 1 m and mass M = 1 kg rotating horizontaly along the vertical axis in its center wit
matrenka [14]

Answer:

w = 0.943 rad / s

Explanation:

For this problem we can use the law of conservation of angular momentum

       

Starting point. With the mouse in the center

            L₀ = I w₀

Where The moment of inertia (I) of a rod that rotates at one end is

         I = 1/3 M L²

Final point. When the mouse is at the end of the rod

          L_{f} = I w + m L² w

As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved

        L₀ = L_{f}

        I w₀ = (I + m L²) w

        w = I / I + m L²) w₀

We substitute the moment of inertia

        w  = 1/3 M L² / (1/3 M + m) L²    w₀

        w = 1 / 3M / (M / 3 + m) w₀

We substitute the values

      w = 1/3 / (1/3 + 0.02) w₀

      w = 0.943 w₀

To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s

        w = 0.943 rad / s

3 0
3 years ago
Which option is an example of reflection?
katovenus [111]

Answer:

A

Explanation:

sound waves hit the canyon walls, then bounce back towards you. That's why you hear an echo

6 0
2 years ago
Read 2 more answers
Please Help Me With This:
yanalaym [24]

To find the relative frequencies, divide each frequency by the total number of throws - in this case, 100.

Red: 10/100=1/10=0.1

Yellow:35/100=0.35

Blue:48/100= 0.48

Misses: 7/100=0.07

The event of landing in the red region has a relative frequency of  0.1 which means the dart landed in the red region about 10% (0.1  x 100%) of the time.

5 0
3 years ago
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