Answer to Question #166201 in Physics for gibson mkandawire

Question #166201

1.Determine the value of a so that vector 𝑨⃗⃗ =(πŸπŸŽπ’ŠΜ‚+πŸπŸ’π’‹Μ‚βˆ’πŸ”π’ŒΜ‚) is perpendicular to vector 𝑨⃗⃗ =(𝟏/πŸπ’ŠΜ‚+𝟏/πŸπ’‹Μ‚+π’‚π’ŒΜ‚)

2. A particle covers a distance of (πŸπ’ŠΜ‚+πŸ‘π’‹Μ‚+πŸ“π’ŒΜ‚) metres in the direction of a constant force of (πŸ•π’ŠΜ‚+πŸ–π’‹Μ‚+πŸ‘π’ŒΜ‚) newtons acting on it. Calculate the work done by the force on this particle.

3. A body moves from position (πŸπ’ŠΜ‚+πŸπŸŽπ’‹Μ‚βˆ’πŸπŸ–π’ŒΜ‚) metres to a position

(πŸπŸ”π’ŠΜ‚+πŸπŸ’π’‹Μ‚+πŸπŸ“π’ŒΜ‚) metres when a force of (πŸ’π’ŠΜ‚+πŸπ’‹Μ‚+πŸ‘π’ŒΜ‚) newton is applied on it. Calculate the work done on a body.

4. A point of application of a force 𝑭⃗⃗ =(πŸπŸ”π’ŠΜ‚+πŸπŸŽπ’‹Μ‚+πŸπŸ“π’ŒΜ‚) is moved from

π’“πŸβƒ—βƒ—βƒ—βƒ— =(πŸπ’ŠΜ‚+πŸ•π’‹Μ‚+πŸπ’ŒΜ‚) to π’“πŸβƒ—βƒ—βƒ—βƒ— =(πŸ“π’ŠΜ‚+πŸπŸπ’‹Μ‚+πŸ‘π’ŒΜ‚). Find the work done.


1
Expert's answer
2021-03-09T15:30:52-0500

1)


"(10,14,-6)\\cdot(0.5,0.5,a)=5+7-6a=0\\\\a=2"

2)


"W=Fr=(2,3,5)\\cdot(7,8,3)\\\\=14+24+15=53\\ J"

3)


"\\Delta r=(16,24,15)-(2,20,-18)=(14,4,33)\\ m"


"W=F\\Delta r=(4,2,3)\\cdot(14,4,33)=163\\ J"

4)


"\\Delta r=(5,12,3)-(2,7,2)=(3,5,1)\\ m"

"W=F\\Delta r=(16,20,15)\\cdot(3,5,1)=163\\ J"


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