Answer:
<u>36 m</u>
Explanation:
We can consider this to be an AP.
Then,
<u>Subtract a₇ from a₉.</u>
- a + 8d - a + 6d = 24 - 20
- 2d = 4
- d = 2
<u>Finding a₁₅</u>
- a₁₅ = a + 14d
- a₁₅ = 8 + 14(2)
- a₁₅ = 8 + 28
- a₁₅ = <u>36 m</u>
Swept-frequency pulses have found use in a variety of fields, including spectroscopic methods where effective spin control is necessary.
To find more, we have to study about the spectroscopic methods.
<h3>
What is homonuclear decoupling and broadband excitation?</h3>
- A thorough understanding of the evolution of spin systems during these pulses is crucial for many of these applications since it not only helps to describe how procedures work but also makes new methodologies possible.
- Broadband inversion, refocusing, and excitation employing these pulses are some of the most popular applications in NMR, ESR, MRI, and in vivo MRS in magnetic resonance spectroscopy.
- A generic expression for chirped pulses will be presented in this study, along with numerical methods for calculating the spin dynamics during chirped pulses using solutions along with extensive examples.
Thus, we can conclude that, the swept-frequency pulses have found use in a variety of fields, including spectroscopic methods where effective spin control is necessary.
Learn more about the broadband excitation here:
brainly.com/question/19204110
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Correct question:
Manny walked a total of 3 miles. The reference point used to calculate the total distance that he walked was the same as the ending point. Which describes where Manny most likely walked?
a. from the bottom of a hill to the top
b. on a circular nature trail
c. on a sidewalk from his house to the mall
d. from the beginning of a straight track to the end
Answer:
b. on a circular nature trail
Explanation:
As it is mentioned that Manny that his reference point from where she started is same as ending point meaning that she moved in a circular path making point B correct answer.
<span>the potential energy of the object.</span>
Answer:
B) 1.5 m/s
Explanation:
The apparent frequency will be enhanced due to Doppler effect
If f be the apparent frequency , F be the real frequency , V be the velocity of sound and v be the velocity of approaching submarine then f is given by
f = F \frac{V+v}{V-v}\\
\frac{f}{F} =\frac{V+v}{V-v}\\
\frac{f}{F}-1 =\frac{V+v}{V-v}-1\\
\Delta f = \frac{2vf}{V-v}\\
200=\frac{2\times v\times 100\times 1000}{1482-v}\\
v=1.48 m/s
