3 years ago

Identifying Mechanical Energy

Physics

2 answers:

sergij07 [2.7K]3 years ago

6 0

Ummmm ok so I don’t really know

arsen [322]3 years ago

3 0

Bruh ima be honest with you idek

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OverLord2011 [107]

Answer:

Part A:

$Displacement\ vector= -40\hat i+20\hat j+25\hat k$

Part B:

Magnitude of displacement=44.721359 m

Explanation:

Given Data:

40 m in negative direction of the x axis

20 m perpendicular path to his left (considering +ve y direction)

25 m up a tower (Considering +ve z direction)

Required:

In unit-vector notation, what is the displacement of the sign from start to end.
The magnitude of the displacement of the sign from start to this new end=?

Solution:

Part A:

$Displacement= (0-40)\hat i+(0+20)\hat j+(0+25)\hat k\\Displacement= -40\hat i+20\hat j+25\hat k$

where:

i,j,k are unit vectors

Part B:

Sign falls to foot of tower so z=0

$Displacement= -40\hat i+20\hat j+0\hat k$

Magnitude of displacement:

$Magnitude\ of\ displacement=\sqrt{(-40)^2+(20)^2+0^2} \\Magnitude\ of\ displacement=44.721359\ m$

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