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mylen [45]
1 year ago
7

The observed a value for the diameter of a Hydrogen atom is 10.1 nm and the accepted value for this is 10 nm. Was you observatio

n accurate?
A. No but it was precise.
B. Yes, the measurement was very close to the actual value.
C. Not enough information.
D. No not at all it was much smaller.
Physics
1 answer:
professor190 [17]1 year ago
3 0

The value of the diameter of the hydrogen atom as we have it in this experiment is accurate because it is very close to the true value.

<h3>What is accuracy?</h3>

The term accuracy refers to how close the measured value is to the true value. The true value is the same as the accepted value. In this case, the true value of the  diameter of a Hydrogen atom is 10 nm.

The fact that is observed value is obtained to be  10.1 nm implies that the value of the diameter of the hydrogen atom as we have it in this experiment is accurate because it is very close to the true value.

Learn more about accuracy:brainly.com/question/14244630

#SPJ1

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A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
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Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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A race car has a mass of 820 kg. It starts from rest and travels 50.0m in 3.0s. The car is uniformly accelerated during the enti
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