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Fittoniya [83]
3 years ago
7

Which action would be most important to perform in a controlled experiment? ___A. Design an experiment that is based on previous

ly published results. ___B. Expose the treated iron to the corrosive effects of the air. ___C. Test a large piece of treated iron and a small piece of untreated iron. ___D. Test a piece of treated iron and a piece of untreated iron under the same conditions.
Physics
1 answer:
ludmilkaskok [199]3 years ago
4 0

Answer:

D. Test a piece of treated iron and a piece of untreated iron under the same conditions.

Explanation:

A controlled experiment is defined as an experiment where the experimenter or the researcher tries to test or experiment his result by changing one variable at a time and keeping the other factors under the same conditions and sees how it effects the result.

The factors which can be changed in an experiment is called its variables.

In the context, testing a piece of treated iron as well as a untreated piece of iron under the same working conditions forms a controlled experiment.

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A sample of chlorine has two naturally occurring isotopes. The isotope Cl-35 (mass 35.0 amu) makes up 75.8% of the sample, and t
irinina [24]

Answer:

M_{av}=35.521amu

Explanation:

As in any sample you will have 75.8% of Cl-35 iosotopes and 24.3% of Cl-37 iosotopes you can get the average atomic mass as:

M_{av}=(35amu*75.8+37amu*24.3)/100=35.521amu

4 0
3 years ago
Three capacitors having capacitances of 8.40, 8.40, and 4.20 μFμF, respectively, are connected in series across a 36.0-V potenti
son4ous [18]

Answer:

a)Q=71.4 μ C

b)ΔV' = 10.2 V

Explanation:

Given that

C ₁= 8.7 μF

C₂ = 8.2 μF

C₃ = 4.1 μF

The potential difference of the battery, ΔV= 34 V

When connected in series

1/C = 1/C ₁ + 1/C₂ + 1/C₃

1/ C= 1/8.4 +1 / 8.4 + 1/4.2

C=2.1 μF

As we know that when capacitor are connected in series then they have same charge,Q

Q= C ΔV

Q= 2.1 x 34 μ C

Q=71.4 μ C

b)

As we know that when capacitor are connected in parallel then they have same voltage difference.

Q'= C' ΔV'

C'= C ₁+C₂+C₃        (For parallel connection)

C'= 8.4 + 8.4 + 4.2 μF

C'=21 μF

Q'= C' ΔV'

Q'=3 Q

3 x 71.4= 21 ΔV'

ΔV' = 10.2 V

3 0
3 years ago
A plastic box has an initial volume of 2.00 m 3 . It is then submerged below the surface of a liquid and its volume decreases to
nikitadnepr [17]

Answer:

Volume strain is 0.02

Explanation:

Volume strain is defined as the change in volume to the original volume.

It is given that,

Initial volume of the plastic box is 2 m³

It is then submerged below the surface of a liquid and its volume decreases to 1.96 m³

We need to find the volume strain on the box. It is defined as the change in volume divided by the original volume. So,

\delta V=\dfrac{V_f-V_i}{V_i}\\\\\delta V=\dfrac{1.96-2}{2}\\\\\delta V=0.02

So, the volume strain on the box is 0.02.

6 0
3 years ago
Someone help please by providing work and answers please :)
Nastasia [14]
First we gotta use an equation of motion:

d = ut + \frac{1}{2} a {t}^{2}

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

100 = 0 + \frac{1}{2} (9.8) {t}^{2}

Now we just need to solve for t:

{t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}

Hit the calculators, and you'll get 4.5 seconds!
5 0
3 years ago
What is the pendulum length whose period is 2.0s ?
Mashutka [201]
Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24mT=2 \pi   \sqrt{\frac{L}{g}} \\
 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi }  = \frac{L}{g} \\
 \frac{T^2}{2 \pi }*g=L\\
L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m

7 0
3 years ago
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