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Paraphin [41]
3 years ago
5

Which of the four cars had the least net force applied to it when it was launched down the ramp

Chemistry
1 answer:
castortr0y [4]3 years ago
5 0

The least net force applied : Car 3(12 N)

<h3>Further explanation  </h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object  

∑F = m. a  

Car 1 ⇒m=0.5 kg, a=36 m/s²

\tt \sum F=0.5\times 36=18~N

Car 2⇒m=0.8 kg, a=50 m/s²

\tt \sum F=0.8\times 50=40~N

Car 3⇒m=0.6, a=20 m/s²

\tt \sum F=0.6\times 20=12~N

Car 4⇒m=1, a=19~m/s²

\tt \sum=1\times 19=19~N

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What is the final pressure of a system (atm) that has the volume increased from 0.75 l to 1.1 l with an initial pressure of 1.25
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Boyle's law states that pressure is inversely proportional to volume of gas at constant temperature 
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The isotope \left.\begin{array}{r}212 \\ 83\end{array}\right? Bi has a half-life of 1.01 yr. What mass (in mg) of a 2.00-mg samp
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Half-life is the length of time it takes for half of the radioactive atoms of a specific radionuclide to decay. A good rule of thumb is that, after seven half-lives, you will have less than one percent of the original amount of radiation.

<h3>What do you mean by half-life?</h3>

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<h3>What affects the half-life of an isotope?</h3>

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5 0
1 year ago
An environmental scientist developed a new analytical method for the determination of cadmium (cd^2+) in mussels. To validate th
Anettt [7]

Answer:

The method is accurate  in the calculation of the Cu^+2

Explanation:

As a first step we have to calculate the <u>average concentration </u>of Cu^+2 find it by the method.

\frac{0.782+0.762+0.825+0.838+0.761 }{5} =0.79 ppm

Then we have to find the<u> standard deviation:</u>

s=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_i-\bar{x})^2}=0.0359

For the confidence interval we have to use the formula:

μ=Average±\frac{t*s}{\sqrt{n} }

Where:

t=t student constant with 95 % of confidence and 5 data=2.78

μ= 0.79  ±  \frac{2.78*0.0359}{\sqrt{5} }

upper limit:  0.84

lower limit: 0.75

If we compare the limits of the value obtanied by the method (Figure 1 Red line) with the reference material (Figure 1 blue line) we can see that the values obtained by the method are within the values suggested by the reference material. So, it's method is accurate.

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