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Vinil7 [7]
3 years ago
7

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

Physics
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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Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc
snow_lady [41]

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force F_g = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force

Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

5 0
3 years ago
A cheetah can run at a maximum speed 103 km/h and a gazelle can run at a maximum speed of 76.5 km/h. If both animals are running
Katena32 [7]

Answer:

1

 t_a  =  13.49 \  s

2

 The distance is   D = 55.2 \  m    

Explanation:

From the question we are told that

   The maximum speed of the cheetah is  v =  103 \  km/h =  28.61 \  m/s

    The maximum of  gazelle is  u =  76.5 \  km/h =  21.25 \  m/s

     The distance ahead is d =  99.3 \  m

Let  t_a denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

           v* t_a = d +  u* t_a

=>          28.61 t_a = 99.3 +  21.25t_a

=>          7.36 t_a = 99.3

=>         t_a  =  13.49 \  s

Now at t =  7.5 s  

             7.5 v = D+  7.5u

=>          28.61 * 7.5  = D +  21.25* 7.5

=>          7.36 *  7.5 =D

=>         D = 55.2 \  m        

Hence the for the gazelle to escape the cheetah it must be 55.2 m

5 0
3 years ago
g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn
krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

            r₁ = 6.780 10⁶ m

for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

            r = 4.21 10⁷ m

this distance is from the center of the earth

7 0
2 years ago
Which substance requires the most heat to melt 1 kg of the substance? A. Copper B. Mercury C. Gold D. Helium
SOVA2 [1]
I would believe it to be C. Gold, but I'm not quite sure
7 0
3 years ago
Which solute will dissolve first in the illustration?
pentagon [3]
B explanation : they are both filled to the same pint
4 0
2 years ago
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