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Alla [95]
2 years ago
8

State advantages of ultrasonic sound in determining the depth of the ocean ​

Physics
1 answer:
agasfer [191]2 years ago
5 0

Answer:

It has <u>greater accuracy than other nondestructive methods in determining the depth of internal flaws and the thickness of parts with parallel surfaces.</u>

Explanation:

Hope this helps you!

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Which source is considered a primary source of scientific information? an original research article in a scientific journal a re
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Newspapers would only pick up the information from the journal itself, or from the authors. Books follow after the original article, after it has gained momentum among the research community. The public lecture at a museum would be based on work from the journal article.
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What is a type of material that tries to prevent the flow of thermal energy?
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A type of material that tries to prevent the flow of thermal energy is called an insulator.
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Please help on this one ?
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D has the largest range
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The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point
makvit [3.9K]
To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
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5 0
3 years ago
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

5 0
3 years ago
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