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2 years ago

Which lists imaging techniques that use wave behaviors in order of resolution from the best resolution to the worst?

Physics

1 answer:

ElenaW [278]2 years ago

3 0

<span>C.CT scan, X-ray imaging, MRI
</span>

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Which free body diagram is in equilibrium?

bearhunter [10]

A. because everything is balanced.

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2 years ago

How would the weight of a 7-kilogram concrete block compared to the weight of a 7-kilogram metal sphere?

Hitman42 [59]

Answer:

C. The block and the sphere would have the same weight.

Explanation:

If both the block and the sphere weigh 7 kilograms then they have the same weight. They may look different, but they both weigh 7 kilograms.

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2 years ago

Read 2 more answers

A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th

Mila [183]

Answer:

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

$\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0$

$\frac{3}{(r + a)^2} = \frac{2}{r^2}$

$\frac{r+a}{r} = \sqrt{\frac{3}{2}}$

$1 + \frac{a}{r} =\sqrt{\frac{3}{2}}$

$\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}$

so we will have

$r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

so the x coordinate of this position is given as

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

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3 years ago

A heat engine operates between two reservoirs at 300K. During each cycle, it absorbs 200 calories from the high temperature rese

yanalaym [24]

(a) Zero

The maximum efficiency (Carnot efficiency) of a heat engine is given by

$\eta=1-\frac{T_C}{T_H}$

where

$T_C$ is the low-temperature reservoir

$T_H$ is the high-temperature reservoir

For the heat engine in the problem, we have:

$T_C = 300K$

$T_H = 300K$

Therefore, the maximum efficiency is

$\eta=1-\frac{300}{300}=0$

(b) Zero

The efficiency of a heat engine can also be rewritten as

$\eta = \frac{W}{Q_H}$

where

W is the work performed by the engine

$Q_H$ is the heat absorbed from the high-temperature reservoir

In this problem, we know

$\eta=0$

Therefore, since the term $Q_H$ cannot be equal to infinity, the numerator of the fraction must be zero as well, which means

W = 0

So the engine cannot perform any work.

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2 years ago

What property of equality can be used to solve the equation. n – 14 = 25

Inga [223]

Do 25-14 and you will get your answer

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2 years ago