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.
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The total distance traveled by Aliaa using her skateboard for 20 revolution of the wheels is equal to 3770 millimeters.
<u>Given the following data:</u>
Diameter of skateboard = 60 mm.
Number of revolution = 20 revolutions.
Radius = diameter/2 = 60/2 = 30 mm.
<h3>What is distance?</h3>
Distance can be defined as the amount of ground covered (traveled) by a physical object over a specific period of time and speed, regardless of its direction, starting point or ending point.
For one revolution of the wheels, the distance traveled by Aliaa using her skateboard is given by:
Distance = 2πr
Distance = 2 × 3.142 × 30
Distance = 188.5 mm.
Therefore, the total distance traveled by Aliaa using her skateboard for 20 revolution of the wheels is given by:
Distance = 188.5 × 20
Distance = 3770 millimeters.
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Answer: 1. h
Explanation:
The block would reach exactly the same height from the ground. It would travel a greater distance away from the source, but the height away from the earth would remain the same as you are giving it the same energy each time. Therefore, it will reach the same gravitation potential energy.
Another approach to look at it this is seeing it when the Block moves up the slope, its kinetic energy decreases and the potential energy increases. In both cases, the kinetic energy decreases by same amount, therefore the block rises to same height H.
Try to use the formula;
1/2MV2 = mgh
Where V = √(2gh)
I hope this helps
Answer:
Consider the followig calculation
Explanation:
a) use deal equation:
PV = nRT
ρ = m/V,= ==> V = m/ρ
therefore,
ρ = Pm/RT
convert 95 oF in degree
95 oF = 308.15 K
1 atm = 1.013 * 105 pascal
ρ = 1.013*105 * 29 * /8.314 * 308.15
= 1.146 kg/m3
b) again use ideal gas equation:
ρ = Pm/RT
T = 50 oF = 283.15 K
1 atm = 1.013 * 105 pascal
molar mass will be same
ρ = 1.013 * 105 * 29 / 8.314 * 283.15
ρ = 1.248 kg / m3
So,
c) . more than density of the hot, dry air computed in part (a)
