All categories

3 years ago

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.

You might be interested in

Which court case provided for Judicial Review?

vfiekz [6]

Answer:

The best-known power of the Supreme Court is judicial review, or the ability of the Court to declare a Legislative or Executive act in violation of the Constitution, is not found within the text of the Constitution itself. The Court established this doctrine in the case of Marbury v. Madison (1803).

6 0

3 years ago

7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has

NNADVOKAT [17]

Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, $\dfrac{m}{t}=30\ kg/s$

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

$F=\dfrac{\Delta P}{\Delta t}$

$F=\dfrac{m(v-u)}{\Delta t}$

$F=30\ kg/s\times (-16-16)\ m/s$

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

6 0

4 years ago

A missle is flying at a speed 125 m/s. if the missle has a mass of 125 kg. What’s it’s kinetic energy

Fynjy0 [20]

Answer:

$976562.5\ Joules$

Explanation:

$We\ are\ given:\\The\ mass\ of\ the\ missile=125\ kg\\The\ velocity\ of\ the\ missile=125\ m/s\\Hence,\\As\ we\ know\ that,\\Kinetic\ energy\ possessed\ by\ an\ object\ with\ mass\ m\ moving\\ with\ a\ velocity\ v:\\E_k=\frac{1}{2}mv^2\\Here,\\The\ Kinetic\ Energy\ possessed\ by\ the\ missile=\frac{1}{2}*125*(125)^2=976562.5\ Joules$

6 0

3 years ago

An object is 12 m long, 0.65 m wide, and 13 cm high. Calculate the volume of this object.

ziro4ka [17]

The volume corresponds to the measure of the space occupied by a body. From the given dimensions we can intuit that we are looking to find the Volume of an Cuboid, that is, an orthogonal rectangular prism, whose faces form straight dihedral angles.

Mathematically the volume of this body is given as

$V = lWh$

Where,

L = Length

W = Width

H = High

$V = (12)(0.65)(13*10^{-2})$

$V = 1.014m^3$

Note: The value given for the height was in centimeters, so it was transformed to meters.

6 0

3 years ago

Residential building codes typically require the use of 12-gauge copper wire (diameter 0.2053 cm) for wiring receptacles. Such c

AysviL [449]

Answer:

a) E = 4.26 W

b) E' = 6.724 W

c) copper wire is the safer option to use.

Explanation:

Given:

Diameter of the 12 gauge copper wire, d = 0.2053 cm

Thus, Radius of the 12 gauge copper wire, r = 0.2053 cm

/ 2 = 0.10265 cm = 0.10265 × 10⁻² m

Now.

the area (A) comes out as

A = π × (0.10265 × 10⁻²)²

A = 3.3103 × 10⁻⁶ m²

Length of the copper wire, L = 2.10 m

a) The resisitivity (ρ) of copper = 1.68 × 10⁻⁸ ohm m

Now,

the resistance of the copper , R = ρL/A

R = (1.68 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

R = 0.01065 ohm

The Energy (E) is given as,

E = I²R

where, I is the current

I = 20.0 A

on substituting the values, we get

E = 20.0² × 0.01065

E = 4.26 W

(b) For the aluminium

Resisitivity, ρ' = 2.65 × 10⁻⁸ ohm m

Now, the resistance of the aluminium wire, R' = (ρ' × L) / A

Since the cross-section of the aluminium wire is same as the copper wire

thus,

R = (2.65 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

R = 0.0168 ohm

Therefore,

The Rate of energy produced by the aluminium wire, E' = I²R'

E' = 20.0² × 0.0168

E' = 6.724 W

(c) From the above results, we can conclude that the power consumed or the rate of energy produced by the aluminium wire is more.

Hence, copper wire is the safer option to use.

8 0

4 years ago