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2 years ago

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

Physics

2 answers:

saul85 [17]2 years ago

5 0

Answer:

I think the answer is C

Explanation:

EDGE 2020

goldenfox [79]2 years ago

3 0

Answer:

C. CT

Explanation:

It stands for Computed Tomography Scans.

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How does a mathematical model help you understand the science concepts

Dmitrij [34]

It can help with measurements and when you want to add measurements to a cylinder or a beaker so ya

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2 years ago

An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and

svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

$V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s$

velocity at the exit=20m/s

for entry

$V=\frac{5}{1} =5m/s$

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

$\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}$

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

$(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81 =p2\\P2=-37.5kPa$

the pressure at exit is -37.5kPa

7 0

3 years ago

What is the peacocks displacement from 2 to 3 seconds

mash [69]

1.5 second difference

7 0

2 years ago

An example of a high energy electromagnetic wave is

Wewaii [24]

<span>An example of a high energy electromagnetic wave is "X-Ray"

When car runs, it's chemical energy (gasoline) converts into mechanical energy

Temperature is the measure of hotness or coldness of the body, so when heat expose to a substance, it's degree of hotness increases & it's temperature increases

Hope this helps!
</span>

4 0

2 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

2 years ago