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hichkok12 [17]
3 years ago
15

Sandra's target heart rate zone is 135bpm—172bpm. Marissa's target heart rate zone is 143bpm—176bpm. They stop playing basketbal

l and take their pulse, and both count the heart rate at 144bpm. If they decrease their heart rates by 20bpm, who will be in her THR zone?
Physics
2 answers:
Feliz [49]3 years ago
6 0

Answer: Neither Sandra nor Marissa will be in her THR zone.


Explanation:


1) Actual pulse of both Sandra and Marissa : 144 bpm


2) Decrease of 20 bpm ⇒ 144 bpm - 20 bpm = 124 bpm


3) Sandra's TRH is in the range 135 - 172 bpm.


Since 124 < 135, she will be below the range.


4) Marissa's TRH range is 143 - 176 bpm.


Since, 124 < 143, she is below the range


In conlusion, neither Sandra nor Marissa will be in her THR zone.

Mekhanik [1.2K]3 years ago
5 0

Sandra's pulse is 144 bpm.  If she decreases it by 20 bpm,
it will be 124 bpm.  Her TRH is the range  of 135 - 172 bpm,
so she will be below it.

Marissa's pulse is 144 bpm.  If she decreases it by 20 bpm,
it will be 124 bpm.  Her TRH is the range  of 143 - 176 bpm,
so she will be below it.

Neither girl will be in her THR.

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J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2

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A = \frac{pL}{R}

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m = dV

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Substitute AL for V in equation of the mass density of copper.

m=d(AL)

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substitute,

pL/R for A  and λ is use for (m/L) in the eqn above

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Substitute 0.200Ω.km⁻¹ for (R/L)

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\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1

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J = \frac{iR}{pL}

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2

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\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m

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