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klemol [59]
3 years ago
12

A man jogs at a velocity of 2.5 m/s west for 1,200 s. He then walks at a velocity 1.0 m/s east for 500 s and then stops to rest

. What is the displacement of man when he stops to rest?
Physics
1 answer:
olganol [36]3 years ago
5 0

Speed of man towards West =2.5 m/s

Time to travel in the given direction = 1200 s

Total displacement towards west

d_1 = v_1 t

d_1 = 2.5 \times 1200 = 3000 m

Now speed of man towards East = 1 m/s

time to travel in east direction = 500 s

total displacement towards East

d_2 = v_2 t

d_2 = 1 \times 500 = 500 m

Now total displacement will be

d = d_1 - d_2

d = 3000 - 500 = 2500 m

so it is 2500 m towards west

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Re_{m} =\frac{p_{m}V_{m}L_{m}   }{u_{m} }

Re_{p} =\frac{p_{p}V_{p}L_{p}   }{u_{p} }

Equaling both Reynold's number:

\frac{p_{p}V_{p}L_{p}   }{u_{p} }=\frac{p_{m}V_{m}L_{m}   }{u_{m} }

Clearing Vm:

V_{m} =\frac{p_{p}V_{p}L_{p} u_{m}   }{u_{p} p_{m} L_{m} }=\frac{999.1*0.56*8*1.849x10^{-5} }{1.138x10^{-3}*1.184*1 } =61.42m/s

b) The drag force is:

\frac{F_{Dm} }{p_{m}V_{m}^{2}L_{m}^{2}     } =\frac{F_{Dp} }{p_{p}V_{p}^{2}L_{p}^{2}     } \\F_{Dp} =\frac{F_{Dp}p_{p}V_{p}^{2}L_{p}^{2} }{p_{m}V_{m}^{2}L_{m}^{2}     } \\F_{Dp}=\frac{2.3*999.1*0.56^{2} *8^{2} }{1.184*61.42^{2}*1^{2}  } =10.32N

6 0
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ioda

Answer:

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