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saw5 [17]
2 years ago
6

You hear the engine roaring on a race car at the starting line. Predict the changes in the sound as the race starts and the car

approaches the camera. A) The sound of the engine will get louder and the pitch lower. Eliminate B) The sound of the engine will get louder and the pitch higher. C) The sound is determined by the car and it will not change in any way. D) The sound of the engine will get louder but the pitch will not change.
Physics
2 answers:
gogolik [260]2 years ago
4 0

B. The sound of the engine will get louder and the pitch higher.

Mariana [72]2 years ago
3 0

Answer: The correct option is B.

Explanation:

Doppler effect is the phenomenon in which there is an apparent change in the frequency when there is relative motion between source and listener.

When the source and the listener are approaching each other then the frequency increases. The loudness depends on the amplitude. The energy of the wave is directly proportional to the square of the amplitude.

In the given problem, the engine is roaring on a race car at the starting line. There is change in the sound as the race starts and the car approaches the camera.

Pitch depends on the frequency.

Therefore, the sound of the wave will get louder and the pitch will get higher.

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Electric current in a solid metal conductor is caused by the movement of what ?
Vlad1618 [11]
Electric current in a solid metal conductor is caused by the movement of electric charge. An electric current is the flow of electric charge. An electron current, the flow of electrons, contributes to an electric current since the electron 'carries' negative electric charge. The flow of ions also contributes to an electric current in, for example, the electrolyte of an electrochemical cell. 
5 0
2 years ago
The movie "The Gods Must Be Crazy" begins with a pilot dropping a bottle out of an airplane. A surprised native below, who think
Sergio039 [100]

Answer:

⇔⇔⇔↑∑∑∩∅¬⊕║⊇↔∴∉∵

Explanation:

8 0
2 years ago
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
A plane increases its speed from 450 km/hr to 750 km/hr in 0.30 hour. What is its acceleration?
ryzh [129]

Answer:

The acceleration is found as:

a = 1000 km/h²

Explanation:

Initial speed of the plane = 450 km/h

Final speed of the plane = 750 km/h

Time taken = 0.3 h

Acceleration can be defined as the change of speed of the object divided by the time taken to bring that change

Acceleration is given as:

a=\frac{v_f-v_i}{t}

where v(f) = 750 km/h , v(i) = 450 km/h , t = 0.3 h

Substitute the values in the equation of acceleration

a= \frac{750-450}{0.3}\\a=\frac{300}{0.3}\\a=1000 km/h^2

This is the found result

5 0
2 years ago
A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro
yanalaym [24]

Answer:

a) v_{o} =16m/s

b) v=9.8m/s

c) \beta =-35.46º

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

x=v_{ox}t

x=v_{o}cos(60)t

v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)} (1)

That would be our first equation

Now, we need to analyze its vertical motion

y=y_{o}+v_{oy}t+\frac{1}{2}gt^2

y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2

Knowing v_{o} in our first equation (1)

8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2

\frac{1}{2}(9.8)t^2=16tan(60)-8

Solving for t

t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

v_{o}=\frac{16m}{(2s)cos(60)}=16m/s

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

v_{x}=v_{ox}+at=16cos(60)=8m/s

v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s

So, the magnitude of the velocity is:

v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s

c) The <u><em>direction of the ball</em></u> is:

\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º

4 0
2 years ago
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