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Butoxors [25]
3 years ago
7

How many significant figures are in this number? 0.00708900

Chemistry
2 answers:
Norma-Jean [14]3 years ago
4 0
Don’t know sorry 1.627
marusya05 [52]3 years ago
3 0
The answer is 6. The two front 0’s after the decimal point are not significant and everything after that is significant which because 1. Non zero numbers are significant
2. Zeros between non zero numbers are significant 3. Trailing zeros are significant (in decimals
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Nataliya [291]
<span>Many scientific investigations have provided evidence to support this as the best explanation of the data</span>
8 0
3 years ago
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What is an experimental control
attashe74 [19]

The experimental control is the standard used as a comparison for the experimental groups.


For example, you may be trying to find out how different types of disinfectants affect bacterial growth.  The control group would receive <em>no</em> disinfectant whereas the experimental groups would be the ones on which the disinfectants were tested.


Hope this makes sense!

3 0
3 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
3 years ago
Find the horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40m/s​
vfiekz [6]

To Find :

The horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40 m/s​.

Solution :

The horizontal range of a projectile is given by :

R = \dfrac{u^2 sin 2\theta}{g} ( Here, g is acceleration due to gravity = 10 m/s² )

Putting all value in above equation :

R = \dfrac{40^2 \times sin (2 \times 15)}{10} \ m\\\\R = \dfrac{1600 \times 1}{2\times 10} \ m\\\\R = 80 \ m

Therefore, the horizontal range of projectile is 80 m.

4 0
3 years ago
Write balanced complete ionic equation for K2SO4(aq)+CaI2(aq)→CaSO4(s)+KI(aq)
tatyana61 [14]
CaSO4(s) might be an improperly capitalized: CAsO4(S), CaSO4(S) Balanced equation: K2SO4(aq) + CaI2(aq) = CaSO4(s) + 2 KI(aq) Reaction type: double replacement.
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