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EXERCISES. Study the two types and complete the table by matching the numbered orthogonal drawings with the same isometric view.

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 EXERCISES. Study the two types and complete the table by matching the numbered orthogonal drawings with the same isometric view. WHAT IS AN ISOMETRIC VIEW AND AN ORTHOGRAPHIC DRAWING? ISOMETRIC VIEW -   is a method for visually representing three-dimensional objects in two dimensions in technical and engineering drawings. ORTHOGRAPHIC DRAWINGS - i s a drawing that shows the top view, front view, and right-side view of a three-dimensional figure.
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Order the steps required to predict the volume of (in mL) of 0.100 M calcium chloride needed to produce 1.00 g of calcium carbonate.

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 Laboratory Experiment: Synthesis of Calcium Chloride 1.) Mixing aqueous solutions of sodium carbonate and calcium chloride to produce solid calcium carbonate. Step 1: Convert the mass of calcium carbonate to moles of calcium carbonate Step 2: Compare moles of calcium carbonate to moles of Calcium chloride based on a balanced equation Step 3: Compute the Volume of calcium chloride solution required Step 4: Convert the volume of calcium chloride solution required 2.) Calculate the Volume (in mL) of 0.100M Calcium Chloride needed to produce 1.00g of Calcium Carbonate (s) Volume of Calcium Chloride = 99.9 mL Solution:
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Three negative point charges lie along a line as shown in the figure. Find the magnitude of the electric field this combination of charges produces at point P, which lies 6.00 cm from the -2.00 μC charge measured perpendicular to the line connecting the three charges. (E = N/C)

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    The problem:   Three negative point charges lie along a line as shown in   F i g .   E 2 1 . 4 1 . Find the magnitude and direction of the electric field this combination of charges produces at point  P P , which lies 6.00 cm from the  − − 2.00- μ μ C charge measured perpendicular to the line connecting the three charges. The Illustration: Given: q1 = q2 = -5.00 μC q3  = -2 μC r1 = 10.00 cm  = 0.1m            r = distance r3 = 6.00 cm = 0.06m Couloumb Constant (k) = 9 ×10 9  kg⋅m 3 ⋅s − 2 ⋅C − 2 The equation to use:  Solution The electric field at P due to q1 and q2 is  E 1 = E 2 = k(q 1 /r1 2 ) = 9x10 9 x (5x10 -6 / (0.1) 2 = 4.5x10 6   N/C The electric field at P due to q3 is E 3 = k(q 3 /r3 2 ) = 9x10 9 x (2x10 -6 / (0.06) 2 = 5x10 6 N/C cos  Θ = 6/10 = 3/5 NET ELECTRIC FIELD E net =  2E 1 (cos Θ ) + E 3         = 2(4.5x10 6 )⅗ ...
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