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3 years ago

9

Joe is an athlete and ran today. His exercise heart rate is 130. His THRZ is 147-179

1 answer:

Katyanochek1 [597]3 years ago

3 0

Answer:

The correct option is;

a) Mike, he is within his target heart rate zone

Explanation:

The Target Heart Rate Zone (THRZ) is the range of the percentage of a person's maximal heart beat rate (MHR) targeted to be achieved during exercising. It is a very good way to gauge the comparative intensity of a person's exercise

Whereby, Joe's heart beat rate is below his THRZ and Mikes heart beat rate is within his THRZ, Mike is working harder.

Hence the correct option is (a) Mike, he is within his target heart rate zone.

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Answer and Explanation:

$Greetings!$

$I'm~Isabelle~Williams~and~I~will~be~answering~your~question!$

$Given:$

$Wave~length~ of~ red ~ light = 683~ nm$

$Wave ~length ~of~ yellow~ light = 576~ nm$

$The~ speed ~of~ the~ Slick~ Willy~ can~ be:$

$\boxed{~f'=(\frac{v+v_{o} }{v} )f~}$

$Where:$

$f'=observed~frequency$

$f=actual~frequency$

$v=speed~of~light$

$v_{o}=~speed~of~observer$

$Plug~in~the~values:$

$683\times 10^-^9=576\times 10^-^9 ~(1+\frac{v_{o} }{3\times 10^8} )$

$\frac{683}{576}=1+\frac{v_{o} }{3\times 10^8}$

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$v_{o}=0.558\times 10^8 ~m/s$

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6 0

4 years ago

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and

puteri [66]

Answer:

14.8 kg

Explanation:

We are given that

$m_1=43.7 kg$

$m_2=12.1 kg$

$g=9.8 m/s^2$

$a=\frac{1}{2}(9.8)=4.9 m/s^2$

We have to find the mass of the pulley.

According to question

$T_2-m_2 g=m_2 a$

$T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N$

$T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N$

Moment of inertia of pulley= $I=\frac{1}{2}Mr^2$

$(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)$

Where $\alpha=\frac{a}{r}$

$(177.87-214.13)=-\frac{1}{2}(4.9)M$

$-36.26=-\frac{1}{2}(4.9)M$

$M=\frac{36.26\times 2}{4.9}=14.8 kg$

Hence, the mass of the pulley=14.8 kg

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