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pishuonlain [190]
3 years ago
14

What kind of degree does an ultrasound tech. need?

Physics
1 answer:
Anna11 [10]3 years ago
7 0
<span>Educational Requirements. Training to become an ultrasound technician can be done through a formal education program or through military training. The most common training is an associate's degree program, although there are bachelor's degree and 1-year certificate programs available.</span>
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A student connects a small solar panel to a 40 a resistor to make a simple circuit. The solar panel produces a voltage of 2 0.
Nastasia [14]
<h3>Solution for the above question : -</h3>

Ohm's law states that :

  • v = ir

the terms used are :

  • r = resistance
  • v = potential \:  \: difference
  • i  = current

let's solve for electric current :

  • 2 = i \times 40

  • i =  \dfrac{2}{40}

  • i = 0.05 \: A

  • i = 50 \: mA

\mathfrak{good\:  \: luck \:  \: for \:  \: your \:  \: assignment}

8 0
2 years ago
A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surf
Sati [7]

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:  

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t_{min = λ/2n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.20

we substitute

t_{min = 750 / 2(1.20)

t_{min = 750 / 2.4

t_{min = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

b)

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t_{min = λ/4n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.50

we substitute

t_{min = 750 / 4(1.50)

t_{min = 750 / 6

t_{min = 125 nm

Therefore, the minimum thickness be now will be 125 nm

4 0
2 years ago
The nuclear power used for electricity is produced by
svet-max [94.6K]

Answer:

<h3>b.fission. </h3>

Explanation:

<h3>Please mark my answer as a brainliest.Please follow me ❤❤❤</h3>
5 0
2 years ago
Read 2 more answers
wite a paragraph explaining the difference between things that have matter and things that don't have matter.
mihalych1998 [28]

Answer: Matter is how much space or opacity an object takes up. In short anything that take up space. Things like balls, trees, and even people are all made of matter. Mass is how much matter a object has. Air also has mass also, even though we can't see it. Things like cars, buildings, even planets have mass.

Explanation: Paragraphs are sometimes 4-6 sentences.

6 0
3 years ago
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

7 0
3 years ago
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