All categories

3 years ago

What is a common use for the physicians Desk Reference PDR

1 answer:

7 0

The PDR is a thick volume that commonly used to provide guidance for prescription drugs available in the United states. The PDR is the most commonly used drug reference . The information is the same as the leaflet found in the drug package also known as the package insert. Drug companies pay to have their drugs included, so many generic medicines will not be found in the book.

For over 60 years, physicians and researchers have turned to PDR for the latest word on prescription drugs. It includes comprehensive and up-to-date information on more than 4000 FDA approved drugs.

You might be interested in

A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is

Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12 m

Let assume the tank have a thickness = dh

The radius of the tank by using the concept of similar triangle is :

$\dfrac{1}{r} = \dfrac{12}{h}$

$r = \dfrac{h}{12}$

The area of the tank = $\mathbf{\pi r^2}$

The area of the tank = $\mathbf{\pi( \dfrac{h}{12})^2}$

The area of the tank = $\mathbf{ \dfrac{\pi}{144}h^2}$

The volume of the tank is = area × thickness

= $\mathbf{ \dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\rho_ g * volume$

where;

$\rho_g$ = density of water ; which is given as 10000 N/m³

So;

Weight of the element = $\mathbf{ 10000 *\dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\mathbf{69.44 \ \pi \ h^2 \ dh}$

However; the work required to pump this water = weight × height rise

where the height rise = 12 - h

the work required to pump this water = $\mathbf{69.44 \ \pi \ h^2 \ dh}$ (12 - h)

the work required to pump this water = $\mathbf{69.44 \pi (12h^2-h^3)dh}$

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = $\mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi[ \frac{12h^3}{3}- \frac{h^4}{4}]^{12}}_0} }$

= $\mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}$

= $\mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}$

= $\mathbf{69.44 \pi*12^4 *\frac{1}{12}}$

= 376966.991 Joules

6 0

3 years ago

Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile? Answ

Vinil7 [7]

Answer: 272.82 drop/tile

Explanation:

Given that the Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile?

Tiles/ft^2 × drop/tiles = drop/ft^2

Tiles will cancel out. Leaving the answer to be drop/ ft^2

Substitutes all the magnitude of the above units.

17 × drop/tiles = 4638

Make drop/tiles the subject of formula

Drop/tiles = 4638/17

Drop/tiles = 272.82

Therefore, 272.82 drop/tile drops fall on each tile?

8 0

3 years ago

Given the diagram showing gas molecules in different containers, a reasonable inference could be made that there is a relationsh

Anvisha [2.4K]

Answer:

The correct answer should be A

Explanation:

4 0

3 years ago

Read 3 more answers

Adding a catalyst to a reaction has an effect similar to

mariarad [96]

The answer to your question is B.

8 0

3 years ago

Write the relation of m² with its a multiple

Harman [31]

Explanation:

The square meter is the SI-derived unit of area. It has a symbol m² (33A1 in Unicode). It is defined as the area of a square whose sides measure exactly one metre.

5 0

3 years ago