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fiasKO [112]
3 years ago
8

please help me ill do anything! Collecting and Analyzing Data There are two types of observations: qualitative and quantitative.

Qualitative observations use words to describe what is observed during the experiment. These notes are kept in a journal or logbook. Quantitative observations use numbers, such as calculations and measurements that are recorded in data tables. Scientists around the world use the same system of measurement called the metric system. In your experiment the unit for volume is cubic meters (m 3 ). The unit for temperature is Kelvin (K). Create a data table for the quantitative data of the lab. Mathematics
Physics
1 answer:
OverLord2011 [107]3 years ago
6 0
It depends on how you want to format your data table, as you can create as many columns or rows as you wish, but the important thing is, to display your quantitative data suitably and in the most simple manner. The main reason why data tables are even used, is to simplify and represent the data in an easy to interpret and access form. If your data table has two columns, then the left side would likely be the manipulated variable, or also known as the independent, and the right side would be the responding or dependent variable. To determine which is which, just look at your data and see which one is manipulated (changed) in order to make the other one "respond" or react to the changes (responding variable). In your case it seems that you do have two units, therefore two "sets" of data, so make sure you know which is which. Also, when you put it into your data table, ensure that you put your units in the top labeling row, so you won't have to rewrite (m 3) for every row of data. After that, it should be fairly easy to input your data and such into the table.
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Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19
Alisiya [41]

Answer:

The workdone is  W = 9.28 * 10^{3} J

Explanation:

From the question we are told that

   The height of the cylinder is  h = 0.588\ m

   The face Area is  A = 4.19 \ m^2

    The density of the cylinder is \rho  =  0.346 * \rho_w

     Where \rho_w is the density of freshwater which has a constant value

              \rho_w = 1000 kg/m^3

     

Now  

     Let the final height of the device under the water be  =  h_f

      Let  the initial volume underwater be = V_n

     Let the initial height under water be  = h_i

      Let the final volume under water be  = V_f

According to the rule of floatation

        The weight of the cylinder =  Upward thrust

This is mathematically represented as

          \rho_c g V_n = \rho_w gV_f

         \rho_c A h = \rho A h_f

So      \frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}

   =>     \frac{h_f}{h_c}  = 0.346

Now the work done is mathematically represented as  

          W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh

               =   \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.

              = \frac{g A \rho}{2}  [h^2 - h_f^2]

              = \frac{g A \rho}{2} (h^2)  [1  - \frac{h_f^2}{h^2} ]

Substituting values

        W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)

        W = 9.28 * 10^{3} J

4 0
3 years ago
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