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Goshia [24]
2 years ago
14

Which of the following statements would be best described as an expression of an attitude that is low in accessibility and high

in ambivalence?
A.
Frank: “I eat a banana for breakfast everyday. I’ve loved them since I was little.”
B.
Judy: “I really dislike bananas. I don’t know why. I just always have.”
C.
Ralph: “I’m not sure about bananas. I’d have to think about it.”
D.
Fiona: “I am a huge banana fan. I don’t know why. I just like them.”
Physics
2 answers:
SpyIntel [72]2 years ago
7 0

Answer:

The answer is C. Ralph: “I’m not sure about bananas. I’d have to think about it.”

Explanation:

I did the practice.

Crazy boy [7]2 years ago
4 0

Answer:

C.

Ralph: “I’m not sure about bananas. I’d have to think about it.”

Explanation:

You might be interested in
Which one of the following represents an acceptable set of quantum numbers for an electron in an atom? (arranged as n, l, m l ,
Vitek1552 [10]

Answer:

The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;

(b) 4, 3, -3, 1/2.

Explanation:

To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;

4, 4, 4, 1/2

4, 3, -3, 1/2

4, 3, 0, 0

4, 5, 7, -1/2

4, 4, -5, 1/2

Let us label them as a to as follows

(a) 4, 4, 4, 1/2

(b) 4, 3, -3, 1/2

(c) 4, 3, 0, 0

(d) 4, 5, 7, -1/2

(e) 4, 4, -5, 1/2

Next we note the rules for the assignment and arrangement of quantum numbers are as follows

Number                                   Symbol                Possible values

Principal Quantum Number  .......n........................1, 2, 3, ......n

Angular momentum quantum

number...............................................l.........................0, 1, 2, .......(n - 1)

Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l  

Spin Quantum Number.................m_s.....................+1/2, -1/2

We are meant to analyze each of the arrangement for acceptability.

Therefore for (a),

we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.

Therefore (a) is not acceptable.

(b) Here we note that

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.

(c) Here we have

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 0 ∉ (+1/2, -1/2) → not acceptable

Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.

(d) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = -1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

(e) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

3 0
2 years ago
An astronaut on a space walk floats a little too far away
Lady bird [3.3K]

Answer:

He can throw it away from himself.

Explanation:

Newtons Third Law says that everything has an equal, yet opposite reaction on other objects.

3 0
1 year ago
Read 2 more answers
Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4
Gre4nikov [31]

Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= (m+ \frac{1}{2} \lambda)

replacing ;

Δx = 2t   ; we have:

2t = (m+ \frac{1}{2} \lambda)

Given that thickness t = 700 nm

Then

2× 700 = (m+ \frac{1}{2} \lambda)     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = (m+ \frac{1}{2} \lambda)     ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness t_{min} ; \ \ \ m =0

∴ 2t_{min} =\frac{\lambda}{2}

t_{min} =\frac{\lambda}{4}

t_{min} =\frac{560}{4}

t_{min} =140 \ \  nm

Thus, the smallest thickness t of the air film = 140 nm

7 0
2 years ago
Read 2 more answers
A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C
mario62 [17]

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}

\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

6 0
2 years ago
The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
1 year ago
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