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azamat
2 years ago
10

How can mechanical waves help in the treatment of cancer?

Physics
2 answers:
Veronika [31]2 years ago
6 0

Beams of X-rays are targeted at a cancerous tumor from outside the body

Explanation:

Mechanical waves helps to treat cancer when beams of x-rays are targeted at a cancerous tumor from outside the body. These beams are energetic and ionizing. This will cause severe destruction of the cancerous cell.

  • In cancer therapy, ionizing radiations are used to destroy cancerous cells in living organisms.
  • Controlled doses of some forms of radiations are particularly known good medical procedures for curing cancer.
  • Other options are not ways in which mechanical waves helps to treat cancer.

learn more:

Lung cancer brainly.com/question/2533779

#learnwithBrainly

Aleks04 [339]2 years ago
4 0

Answer:

The answer would be D

Explanation:

Just did it on ed

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What is the explanation for how a modern transmission electron microscope (TEM) can achieve a resolution of about 0.2 nanometers
IgorC [24]

Answer:

Explanation:

A simple light microscope uses light for imaging of objects where as a transmission electron microscope uses a monochromatic beam of electrons.

This beam is passed by a magnetic field which is very strong and thus act as a lens.

Its resolution of very high which is about 0.2 nanometers because of the separation between two atoms.

Because of this reason its resolution is about 1000 times greater than light microscope.

3 0
2 years ago
Which statement correctly describes the relationship between frequency and wavelength?
Len [333]
The relationship between the frequency and wavelength of a wave is given by the equation:

v=λf, where v is the velocity of the wave, λ is the wavelength and f is the frequency. 

If we divide the equation by f we get:

λ=v/f

From here we see that the wavelength and frequency are inversely proportional. So as the frequency increases the wavelength decreases. 

So the second statement is true: As the frequency of a wave increases, the shorter the wavelength is.  
3 0
2 years ago
Read 2 more answers
Suppose an X-ray binary is found in which the visible star is a 12 M⦿ red giant, the orbital period is 3.65 days, and the semima
nataly862011 [7]

Answer:

If the mass of a star is greater than 3 solar masses, it will create a black hole. If its mass is less, it will create a neutron star.

Explanation:

If a star's gravity is high enough, when it condenses on itself, it will form a black hole. Otherwise, it will create a large amount of highly dense matter, such as a neutron star. It can be said that if the mass of a star is greater than 3 solar masses, it will create a black hole. If its mass is less, it will create a neutron star.

8 0
2 years ago
Corey, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 850 N for the first 3.0 s after the eleva
lyudmila [28]

Answer:

v₂ = 2.568 m/s

Explanation:

given,

mass of Corey = 95 Kg

reading of sale for first 3 s when elevator start to move = 850 N

scale reading for the next 3.0 s = 930 N

Gravitation force acting =

  F = m g

  F = 95 x 9.8

  F = 931 N

using newtons second law, due to movement of elevator

    F_{net} = m a

 W - N = m a₁

931- 850 = 95 x a₁

    a₁ = 0.852 m/s²

now,

velocity calculation

v₁ = a₁t

v₁ = 0.852 x 3 = 2.557 m/s

now, For second case

931 - 930 = 95 x a₂

a₂ = 0.011 m/s²

now, velocity after 4 s

v₂ = v₁ + a₂ t

v₂ = 2.557+ 0.011 x (4 - 3)

(4-3) because velocity after 3 second is calculate we need to calculate velocity after 4 s from beginning.

v₂ = 2.557 + 0.011

v₂ = 2.568 m/s

velocity of the elevator is equal to v₂ = 2.568 m/s

7 0
2 years ago
An airplane weighing 5000 lb is flying at standard sea level with a velocity of 200 mi/h. At this velocity the L/D ratio is a ma
saul85 [17]

Answer:

98.15 lb

Explanation:

weight of plane (W) = 5,000 lb

velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s

wing area (A) = 200 ft^{2}

aspect ratio (AR) = 8.5

Oswald efficiency factor (E) = 0.93

density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}

Drag = 0.5 x ρ x v^{2} x A x Cd

we need to get the drag coefficient (Cd) before we can solve for the drag

Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)

where

  • induced drag coefficient (Cdi) = \frac{Cl^{2} }{n.E.AR} (take note that π is shown as n and ρ is shown as p)    

        where lift coefficient (Cl)= \frac{2W}{pAv^{2} }=\frac{2x5000}{0.002377x200x293.3^{2} } = 0.245

        therefore

       induced drag coefficient (Cdi) = \frac{Cl^{2} }{n.E.AR} = \frac{0.245^{2} }{3.14x0.93x8.5} = 0.0024

  • since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)
  • Cd = 0.0024 + 0.0024 = 0.0048

Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.        

 Drag = 0.5 x ρ x v^{2} x A x Cd

Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb

8 0
2 years ago
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