A student applies 0.3 N of force to a computer mouse. The mouse slides .025 m. How much work did the student do on the mouse?
1 answer:
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The cost of one roll of plain wrapping paper is $5 and one roll of shiny wrapping paper costs $16.
Step-by-step explanation:
Step 1; Assume that the cost of one roll of plain wrapping paper is $x and a roll of shiny wrapping paper costs $y. so we have the following equations from the given data
4x + 3y = 68, take this as equation 1.
9x + 2y = 77, take this as equation 2
Step 2; We multiply equation 1 with 2 and equation 2 with 3 so we can cancel out the variable y in both equations. By doing this we get
8x + 6y = 136, take this as equation 3,
9x + 2y = 77, this is equation 4.
If we subtract 4 from 3, we cancel out the y variable and can calculate the value of x.
-19x = -103 , x = = 5.
Step 3; Substituting this value of x in any of the previous equations we will get x's value. Here this value of y is substituted in equation 1.
4 (5) + 3y = 68, 20 + 3y = 68 , 3y = 48, y= 16.
So we have x = $5 and y = $16.