The answer is =87/126〈9,6,3〉 The vector projection of vecb onto veca is = (veca.vecb)/ (∥veca∥^2)veca The dot product is veca.vecb=〈8,2,1〉.〈9,6,3〉 = (72+12+3)=87 The modulus of veca is …

Explanation: #"Step 1 is calculating the modulus of the number."# #| -6 + i | = sqrt (6^2 + 1) = sqrt (37)# #"Step 2 is multiplying and dividing through that modulus ...

DeltaL=8.88296m... DeltaL=8882.9mm Use F/A=Y(DeltaL)/(L_0 which relates Youngs Modulus with Tensile Stress and Tensile Strength . Since diameter equals 2.7 mm, r=1.3*10^-3m . Use A=pir^2 …

The result is =-183.9 The dot product is vecA.vecB=<8,-2.9>.<5,4,-9> = (8*5)+ (-2*4)+ (9*-9) =40-8-81=-49 The modulus of vecA is ||vecA||=||<8,-2,9>||=sqrt (64+4+81 ...

The angle is =42.7^@ Let's start by calculating vecC=vecA-vecB vecC=〈4,-5,1〉-〈5,8,-3〉=〈-1,-13,4〉 The angle between vecA and vecC is given by the dot product definition. …

see explanation First do the normal graph of y=6x-7 without the modulus. graph {6x-7 [-10, 10, -5, 5]} Since the modulus is only applied to the x, think of it as a function. It is inside the bracket so you shift …

The angle is =62.6º We start by calculating vecC vecC=vecA-vecB =〈7,-5,6〉-〈4,4,9〉=〈3,-9,-3〉 The angle between 2 vectors is given by the dot product. vecA ...

"magnitude=47.17 units , angle=122.01"^o "let the x-component of the vector be "F_x" (it has shown as green)." "let the y-component of the vector be "F_y" (it has shown as blue)." "The F vector shows the …

The two are related by stiffness (or “Young modulus” if we’re going to be posh) as the stiffness is the stress divided by the strain. This too has units of Pascals.

The angle is =56º Let's start by calculating vecC=vecA-vecB vecC=〈5,2,-5〉-〈6,5,9〉=〈-1,-3,-14〉 The angle between vecA and vecC is given by the dot product definition. …