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Inga [223]
3 years ago
7

Computer-enhanced x-rays are called __________ scans. A. TMS B. PET C. CT D. MRI

Physics
2 answers:
saul85 [17]3 years ago
5 0

Answer:

I think the answer is C

Explanation:

EDGE 2020

goldenfox [79]3 years ago
3 0

Answer:

C. CT

Explanation:

It stands for Computed Tomography Scans.

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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
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Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
3 years ago
How long does it take for a Ford Econoline van moving at 39.5 m/s to travel 600 m?
Andrej [43]

Answer: B. 15.2s

Explanation: 600/39.5 = about 15.2

6 0
4 years ago
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Annette [7]

In the beginning, the melting stage they call it.

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Vikentia [17]
B) using a blood pressure monitor to find a patient's blood pressure.

Hope this helped!
8 0
3 years ago
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