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Ilia_Sergeevich [38]
3 years ago
8

Research is being done on the use of radio waves in destroying cancer cells. What type of frequency would be best used in this t

echnology? Justify your answer.
Physics
2 answers:
Dima020 [189]3 years ago
5 0
This is a <span>John Kanzius Research. </span>Radiofrequency Ablation (RFA) which is a type of Electrical Energy is used to treat cancer. It u<span>ses heat made by </span>radio waves<span> to kill </span>cancer<span> cells. Ablation means destroying completely. The method is putting RFA  a probe (electrode) that goes through the skin into the tumour. A</span> CT scan is needed so to check that the probe is in exactly the right place. An electrode in the probe creates radiofrequency energy to produce heat and kill the cancer cells. 
IRISSAK [1]3 years ago
5 0

Destroying cancer cells requires a lot of energy.

High-frequency waves have high energy.

--

The destruction of cancer cells requires a lot of energy, so high-frequency radio waves can be used in this technology because they have high energy.

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What caused magnetic fields? What happens If you keep breaking a magnet in two until you reach a single atom?
mote1985 [20]

magnetic materials will produce magnetic field near it

All magnets are made up of small magnet type atoms which are known as domains

These all atoms will align itself so that they all produce strong magnetic field along the axis

this magnetic field will become more strong as more number of atoms are aligned.

Now if we break the magnet into small piece then the magnetic field strength will start decreasing and at the end when only one atom will remain then the magnetic field strength will reduce to the field of one atom only.

6 0
3 years ago
Vectors A and B lie in the xy-plane. Vector A has a magnitude of 19.1 and is at an angle of 125.5º counterclockwise from the +x-
Nady [450]

Answer:

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Explanation:

l

8 0
3 years ago
A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp
Jobisdone [24]

Answer:

The tension is  T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by hinge   Fx= \frac{11}{4\sqrt{3} } Mg

Explanation:

   From the question we are told that

          The mass of the beam  is   m_b =M

          The length of the beam is  l = L

           The hanging mass is  m_h = 3M

            The length of the hannging mass is l_h = \frac{3}{4} l

            The angle the cable makes with the wall is \theta = 60^o

The free body diagram of this setup is shown on the first uploaded image

The force F_x \ \ and \ \ F_y are the forces experienced by the beam due to the hinges

      Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

     So

           \sum F =0

Now about the x-axis the moment is

              F_x -T cos \theta  = 0

     =>     F_x = Tcos \theta

Substituting values

            F_x =T cos (60)

                 F_x= \frac{T}{2} ---(1)

Now about the y-axis the moment is  

           F_y  + Tsin \theta  = M *g + 3M *g ----(2)

Now the torque on the system is zero because their is no rotation  

   So  the torque above point 0 is

          M* g * \frac{L}{2}  + 3M * g \frac{3L}{2} - T sin(60) * L = 0

            \frac{Mg}{2} + \frac{9 Mg}{4} -  T * \frac{\sqrt{3} }{2}    = 0

               \frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}

               T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }

                   T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by the hinge is

             F_x= \frac{T}{2} ---(1)

Now substituting for T

              F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}

                  Fx= \frac{11}{4\sqrt{3} } Mg

4 0
3 years ago
7. You probably had difficulty in measuring the period, yet its calculation should prove effortless at this point. Why do you su
Art [367]

Answer:

* First fight the reaction time of the person, which is approximately 1/10 s, remember that the stopwatch does not stop alone

* Problems to synchronize when starting to measure the oscillation and end point of the oscilloscope, this error needs a good control system to be decreased.

*Effect of friction with air

Explanation:

In the measurements of the oscillatory movement in general, the most difficult magnitude of the measurement is the period, by reasoning

* First fight the reaction time of the person, which is approximately 1/10 s, remember that the stopwatch does not stop alone

* Problems to synchronize when starting to measure the oscillation and end point of the oscilloscope, this error needs a good control system to be decreased.

* Effect of friction with air

5 0
3 years ago
A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of
Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.  

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

K_1 = K_2 + P + W

then, the work done by the engine is:

W = K_1 - K_2 - P

The formulas for the Kinetic and potential energy are:  

K=\frac{1}{2}mV^2\\P=mgh

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh

Replacing the values:

W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

t=\frac{3200m}{20m/s}=160s

the power delivered by the car's engine was:

power=\frac{work}{time}=\frac{1690400J}{160s}=10565W

8 0
4 years ago
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