Recent content by farhana21

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    Vectors using the component method

    Please see this thread for figure https://www.physicsforums.com/showthread.php?t=670846
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    Vectors using the component method

    Use the component method to add the vectors and shown in the figure. The length of is 2.55 m and the angle θ = 31.5°. Express the resultant + in unit-vector notation. The answer should be in the form of ...i + ...j so far i have done 3*cos(31.5) = 2.56 so this is for ...i but I am...
  3. F

    Vectors using the component method

    Use the component method to add the vectors and shown in the figure. The length of is 2.55 m and the angle θ = 31.5°. Express the resultant + in unit-vector notation. The answer should be in the form of ...i + ...j so far i have done 3*cos(31.5) = 2.56 so this is for ...i but I...
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    Physics Wordy Question: Calculating Areas and Masses on a Graph - Help Needed!

    For part (c) I chose a grid point on the line far from the origin: slope = 0.31 g/600 cm2 = 0.000 52 For part (b) I did (200, 0.1) (300, 0.15). Then (0.15-0.1)/(300-200) = 5* 10^-4 * 100 *100 = 5%
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    Physics Wordy Question: Calculating Areas and Masses on a Graph - Help Needed!

    The answer for part c is The answer to part c is 0.00052 g/cm^3 would someone please kindly tell me the answer to part b. im really stuck. help is most appreciated Thank you
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    Physics Wordy Question: Calculating Areas and Masses on a Graph - Help Needed!

    Can someone please help me on this question please. A student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. He cuts out various shapes in various sizes, calculates their areas, measures their masses and prepares the graph in the figure below...
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    How Many Copper Atoms Are in a Cubic Centimeter?

    Can someone please help me on this physics question The mass of a copper atom is 2.50 10-25 kg, and the density of copper is 8 920 kg/m3 . (a) Determine the number of atoms in 1 cm3 of copper. (b) Visualize the one cubic centimeter as formed by stacking up identical cubes, with one...
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