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3 years ago

7

Which property best makes radio waves safe for diagnosing illnesses through magnetic resonance imaging?

2 answers:

77julia77 [94]3 years ago

8 0

Answer:

1) Long energy of radio waves

2) long wavelengths of radio waves

Explanation:

aleksandrvk [35]3 years ago

4 0

Answer:

It is used in MRI because it does not damage cells

Radio waves are used for space research because they have very long wavelengths

Explanation:

Many parts of the electromagnetic spectrum are applied in clinical diagnosis and treatment of illnesses. However, these highly ionizing radiation damage cells and its dosage must be carefully managed to avoid creating radiation related health problems for the patients.

Radio waves can be used in MRI without issues because the energy of the radiation is not sufficient to cause damage to cells but is sufficient to provide images for the sake of medical diagnosis.

Secondly, radio waves have long wavelength. This property is suitable for long range

communication. Hence it can be used in space research

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Coulomb's law states that the force F of attraction between two oppositely charged particles varies jointly as the magnitude of

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Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved

Explanation:

Let the charges be q1 and q2 and the distance between the charges be 'd'

Mathematical representation of coulombs law will be;

F1=kq1q2/d²...(1)

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If q1 and q2 is doubled and the distance halved, we will have;

F2 = k(2q1)(2q2)/(d/2)²

F2 = 4kq1q2/(d²/4)

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F1/F2 = kq1q2/d² ÷ 16kq1q2/d²

F1/F2 = kq1q2/d² × d²/16kq1q2

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A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released

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Answer:

$6.75\mu C/m^2$

Explanation:

We are given that

Diameter,d= $1\mu m=1\time 10^{-6} m$

$1\mu m=10^{-6} m$

Radius,r= $\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m$

Density, $\rho=900kg/m^3$

Total number of electrons,n=39

Charge on electron = $1.6\times 10^{-19} C$

Total charge= $q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C$

Distance,s=2mm= $2\times 10^{-3} m$

Mass = $density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg$

Initial velocity,u=0

Final speed,v=4.5 m/s

$v^2-u^2=2as$

$(4.5)^2-0=2a(2\times 10^{-3})$

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$a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2$

Force,F=ma

$qE=ma$

$q(\frac{\sigma}{2\epsilon_0})=ma$

$\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}$

$\epsilon_0=8.85\times 10^{-12}$

$\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2$

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3 years ago