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2 years ago

A beaker contain 200mL of water What is its volume in cm3 and m3

Physics

2 answers:

Bogdan [553]2 years ago

7 0

The volumes are 200cm3 and 0.0002m3

Nutka1998 [239]2 years ago

4 0

A mL is equal to a cm^3, so it is 200 cm^3. To help you transfer cm^3 to m^3, use KHDBDCM. You move the decimal point twice to the left which means that it is 2 m^3.
TLDR: 200 cm^3 and 2 m^3.

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Bradley gets an x-ray at a radiology clinic that employs its own technologists and radiologists. Would the coder at the clinic r

KengaRu [80]

Answer:

Explanation:

If Bradley examination was done and interpreted in the same facility, the radiologist code is used example- procedure code 72100- Radiologic examination, spine, lumbosacral, 2 or 3 views is reported.

if the X-ray was taken by Dr X but Dr X does not read or interpret the image but forward it to the radiologist for initial report, then a 26- modifier is used. E.g A reports by the technologist would be, procedure code 72050-Radiologic examination, spine, cervical, 2 or 3

views or 72050- TC in certain situations and the consulting radiologist would report 72050-26.

if Bradley’s x-ray were sent to an independent radiologist for interpretation, then the procedure code 76140 is used in reporting.

8 0

1 year ago

A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is

Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

$F = AY\alpha \Delta T$

Where

A = Cross-sectional Area

Y = Young's modulus

$\alpha$ = Coefficient of linear expansion for steel

$\Delta T$ = Temperature Raise

Our values are given as,

$A = 4.45cm^2$

$T = 37K$

$\alpha = 1.17*10^{-5}K^{-1}$

$Y = 200*10^9Gpa$

Replacing we have,

$F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)$

$F = 38526.1N$

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

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2 years ago

A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a

vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

$a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2$

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So $gsin\alpha = 4.43cos\alpha$

$\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451$

$tan\alpha = 0.451$

$\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0$

3 0

2 years ago

A 29 cm pencil is placed 35cm in front of a convex lens and is illuminated by a spotlight. the focal point of the lens is 28cm f

vovikov84 [41]

A) What is the height of the pencil image

4 0

2 years ago

A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the

Alborosie

Answer:

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = $\frac{\pi d^{2} }{4}$

where d is the diameter of the wire

1.613 x 10^-7 = $\frac{3.142* d^{2} }{4}$

6.448 x 10^-7 = 3.142 x $d^{2}$

$d^{2}$ = $\sqrt{ 2.05*10^-7}$

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0

2 years ago