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Rufina [12.5K]
3 years ago
11

With a magnetic field strength of 1.41 tesla, all of the protons in organic compounds will resonate over a narrow range of frequ

encies near 60 MHz. With a magnetic field strength of 7.04 tesla is employed, all of the protons in organic compounds will resonate over a range of frequencies near
Physics
1 answer:
juin [17]3 years ago
7 0

Answer:

<h3>300 MHz</h3>

Explanation:

A resonance is obtained when oscillating frequency of magnetic field becomes approximately equal to intrinsic frequency of particles in the nuclei

this method is used in Nuclear Magnetic Resonance or NMR

here the resonant frequency is <em>directly proportional</em> to the strength of magnetic field

from the question,

   initial Magnetic Field = 1.41 Tesla

   Initial resonant frequency = 60 MHz

   final Magnetic Field = 7.04 Tesla

   final resonant frequency = \frac{7.04}{1.41}  X 60 = 5 X 60 = 300 MHz

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A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9
Andreyy89

Answer:

d = 0.0306 m

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

U + K = constant

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0

7.05 + 30.5 = \frac{0.266}{r}

r = 7.08 \times 10^{-3} m

so distance moved by the particle is given as

d = r_1 - r_2

d = 0.0377 - 0.00708

d = 0.0306 m

6 0
3 years ago
A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh
Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

F_s = F_o [\frac{v}{v_s + v} ] \\\\

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0
3 years ago
What is the purpose of the wire coil in an electromagnet?
ziro4ka [17]
It is to conduct electricity in the magnet so it has an electric field.

Please BRAINLIEST!
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3 years ago
sandra lives in a country located in the northern hemisphere she makes a sundial by erecting a pole vertically in her garden the
Arte-miy333 [17]

Answer:

C

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Not really but I need points lol
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