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Sloan [31]
3 years ago
13

WHICH SHOULD NOT BE PART OF SCIENTIFIC INQUIRY

Physics
1 answer:
umka21 [38]3 years ago
5 0

Answer:

Not collecting the info

Explanation:

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Answer: Chemical

Explanation: because it is changeing its chemical structure if its wrong sorry .

3 0
3 years ago
I WILL MARK BRAINLIEST TO WHOEVER GETS THIS CORRECT
Veseljchak [2.6K]

Answer:

If south-east Texas is where H then it is C, but if not the answer is A.

6 0
2 years ago
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
Please help me with this question guys.
katen-ka-za [31]

Answer:

<em>The average speed is 22.2 km/h</em>

Explanation:

<u>Average Speed</u>

Given an object travels a total distance d and took a total time t, then the average speed is:

\displaystyle \bar v=\frac{d}{t}

The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:

\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h

Then he drives d2=7 km at v2=43 km/h taking a time of:

\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h

The total time is

t=0.467 h + 0.163 h = 0.63 h

The total distance is

d = 7 km + 7 km = 14 km

The average speed is:

\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h

The average speed is 22.2 km/h

7 0
2 years ago
Define rotational speed?- short answer
Marina86 [1]
Rotational speed of an object around an axis is the numbers of turns of the object divided by time.  <span />
3 0
3 years ago
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