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babymother [125]
2 years ago
8

Two red and two overlapping balls in the center are surrounded by a green, fuzzy, circular cloud with a white line running throu

gh it. Two green balls sit on the white line, and a line leading a bracket around the balls is labeled A. A line leading to a bracket overlapping the white line is labeled B.
Physics
2 answers:
PolarNik [594]2 years ago
3 0

Answer:

  • A. The line leading a bracket around the balls, labeled A, represents the nucleus of the atom.

  • B. The white line, labeled B, is signaling the first main energy level, as per the quantum model of the atom.

Explanation:

You are describing the model of an atom with two protons, two neutrons, the electron density (orbital), and two electrons.

Then, you want to know what the <em>labels A</em> and <em>B</em> represent.

The nucleus of the atom is at the center, where protons and neutrons are placed. This is supported by three models: Rutherford's model, Bohr's model, and the (currently accepted) quantum model.

Hence, the red balls and the two overlapping balls in the center are representing the protons and neutrons.

The <em>line leading a bracket around the balls</em>, labeled A, represents the nucleus of the atom, since it is signaling both protons and neutrons in the nucleus.

The green fuzzy circular cloud depicts the region where the electrons are: it is fuzzy in correspondence to the quantum model, which states the electrons are not in a fixed position but in a region around the atom. That region is named orbitals. As per the model you cannot tell the exact position of the electron, and that is what the fuzzy cloud means.

In the the Bohr's model that white line,  which should be circular, where the electrons are, depicts the orbit of the electrons, and the shell, which were idenfified with letters L, M, N, K. The first shell was the L shell.

In the quantum  model, that line corresponds to the principal quantum number n, which is the main energy level. Hence, the line leading to a bracket overlapping the white line depicts the main energy level.

The atom is neutral because it contains the same number of electrons aroud the nucleus as protons inside the nucleus.

dem82 [27]2 years ago
3 0

Answer:

Explanation:Which statement describes an atomic nucleus?

An atomic nucleus is negatively charged.

An atomic nucleus has no electrical charge.

An atomic nucleus consists of mostly empty space.

An atomic nucleus contains most of the atom’s mass.

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(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is
Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H  = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = \dfrac{1}{2}mv^2

Using Conservation of energy

KE_i + PE_i + KE_f + PE_f

0 + m g H = \dfrac{1}{2}mv^2 + 0

v = \sqrt{2gH}

v = \sqrt{2\times 9.8 \times 105}

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0
2 years ago
Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at
Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

v = \dfrac{2\pi R}{T}

v = \dfrac{2\pi\times 8}{4.30}

  v = 11.69 m/s

now, Force does the ring push on her at the top

- N - m g = \dfrac{-mv^2}{R}

N + m g = \dfrac{mv^2}{R}

N = \dfrac{mv^2}{R}- m g

N = m(\dfrac{v^2}{R}- g)

N = 58\times (\dfrac{11.69^2}{8}- 9.8)

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0
2 years ago
Assume a rectangular strip of a material with an electron density of n=5.8x1020 cm-3. The strip is 8 mm wide and 0.8 mm thick an
vampirchik [111]

Answer: I = 111.69 pA

Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.

The hall voltage of a semiconductor sensor is given below as

V = I×B/qnd

Where V = hall voltage = 1.5mV =1.5/1000=0.0015V

I = current =?,

n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3

q = magnitude of an electronic charge=1.609×10^-19c

B = strength of magnetic field = 5T

d = thickness of sensor = 0.8mm = 0.0008m

By slotting in the parameters, we have that

0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008

0.0015 = I×5/7.446×10^-8

I = (0.0015 × 7.446×10^-8)/5

I = 111.69*10^(-12)

I = 111.69 pA

3 0
3 years ago
The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad
kenny6666 [7]

Answer:

a)  v = 2,9992 10⁸ m / s , b)  Eo = 375 V / m ,  B = 1.25 10⁻⁶ T,

c)     λ = 3,157 10⁻⁷ m,   f = 9.50 10¹⁴ Hz ,  T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

        E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

        v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

        k = 2π / λ

        λ = 2π / k

        λ = 2π / 1.99 10⁷

        λ = 3,157 10⁻⁷ m

        w = 2π f

        f = w / 2 π

        f = 5.97 10¹⁵ / 2π

        f = 9.50 10¹⁴ Hz

the wave speed is

        v = 3,157 10⁻⁷   9.50 10¹⁴

        v = 2,9992 10⁸ m / s

b) The electric field is

           Eo = 375 V / m

to find the magnetic field we use

           E / B = c

           B = E / c

            B = 375 / 2,9992 10⁸

            B = 1.25 10⁻⁶ T

c) The period is

           T = 1 / f

            T = 1 / 9.50 10¹⁴

            T = 1.05 10⁻¹⁵ s

the wavelength value is

          λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

this wavelength corresponds to the ultraviolet

5 0
3 years ago
Rita jeptoo of kenya was the first female finisher in the 110th boston marathon. she ran the first 10.0 km in a time of 0.5689 h
stira [4]

Part a

Answer: 17.58 km/h

Average speed=\frac{Total\hspace{1mm}Distance}{Total\hspace{1mm}Time}

Total Distance =10 km

Total time =0.5689 h

\Rightarrow Average speed=\frac{10\hspace{1mm}km}{0.5689\hspace{1mm}h}=17.6 \hspace{1mm}km/h

Part b

Answer: 17.626 km/h

Average speed=\frac{Total\hspace{1mm}Distance}{Total\hspace{1mm}Time}

Total Distance =42.195 km

Total time =2.3939 h

\Rightarrow Average speed=\frac{42.195\hspace{1mm}km}{2.3939\hspace{1mm}h}=17.626\hspace{1mm}km/h

8 0
2 years ago
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