Explain how rays AB and AC form both a line and an angle.

Response:

A 120 cm long pipe produces resonance at wavelengths of 480 cm, 160 cm, and 96 cm, but fails to resonate with any wavelengths longer than those. Consequently, this pipe is classified as:

A. closed at both ends

B. open at one end and closed on the other

C. open at both ends.

D. we cannot determine since we lack information about the sound frequency.

The correct answer is:

B. open at one end and closed at the other.

Explanation in steps:

The given data states that the pipe has a length of

= 120 cm

with wavelengths = 480 cm

                             

= 160 cm and = 96 cm

We need to ascertain if the pipe is open, closed, or of an open-closed nature.

Note:

For a pipe open at both ends, the fundamental wavelength is 2L.

For a pipe closed at one end and open at the other, the fundamental wavelength is 4L.

• From this information,
• It appears that this pipe is open at one end and closed at the other.

Now let’s verify with the subsequent wavelengths.

In the case of a pipe with one open end and one closed end:[

Odd multiples of one-quarter wavelength must fit within the length L.

⇒  

                           ⇒  

⇒

                               ⇒  

⇒                                    ⇒  ⇒

                             ⇒  

 Thus, it confirms that the pipe is open at one end and closed at the other.

Please clarify the question as I don't quite understand it.

A) The speed of pedestrian BC is 5 km/h

    The speed of pedestrian CD is 0 km/h

    The speed of pedestrian DE is 5 km/h

B) He reached E since the stop after 6 hours

C) The formula for section BC is d(t) = 40 - 5t

    The formula for section CD is d(t) = 20

    The formula for section DE is d(t) = 50 - 5t

Step-by-step breakdown:

A)

In the time-distance graph, speed represents the change in distance with respect to time

this means speed = Δd/Δt ⇒ (slope of the line)

For line BC:

1. Δd = 40 - 20 = 20 km

2. Δt = 4 - 0 = 4 hours

3. The speed = 20 ÷ 4 = 5 km/h

The speed of pedestrian BC stands at 5 km/h

For line CD:

1. Δd = 20 - 20 = 0 km

2. Δt = 6 - 4 = 2 hours

3. The speed = 0 ÷ 2 = 0 km/h

The speed of pedestrian CD is 0 km/h

For line DE:

1. Δd = 20 - 0 = 20 km

2. Δt = 10 - 6 = 4 hours

3. The speed = 20 ÷ 4 = 5 km/h

The speed for pedestrian DE is 5 km/h

B)

∵ He stopped at t = 4 hours

∵ He reached point E at t = 10 hours

∵ 10 - 4 = 6 hours

He arrived at E since the stop took 6 hours

C)

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1. f(x) is referenced as d(t)

2. m is the speed

3. x corresponds to t

4. You can calculate c by substituting any coordinates of a point along the line into the formula

<pline bc="">

Line BC has a negative slope because d decreases as t increases

∵ m = -5 and c = 40

Thus, d(t) = 40 - 5t

The equation for section BC is d(t) = 40 - 5t

Line CD

Line CD is a horizontal line (which follows the form of any horizontal line represented as y = c)

Thus, m = 0 and c = 20

Therefore, d(t) = 20

The equation for section CD is d(t) = 20

Line DE

Line DE features a negative slope as d decreases along with an increase in t

∵ m = -5

Thus, d(t) = -5t + c

The value of c can be determined by substituting point D's coordinates into the equation

∵ D's coordinates are (6, 20)

Thus, 20 = -5(6) + c

So, 20 = -30 + c

Adding 30 to both sides yields

∴ c = 50

Thus, d(t) = 50 - 5t

The equation for section DE is d(t) = 50 - 5t

Further Learning:

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