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Sever21 [200]
3 years ago
13

2)Prueba de presión Cuando a una persona se le somete a una prueba de presión, por lo general se le indica que, al llegar el rit

mo cardíaco a cierto punto, la prueba deberá detenerse. El máximo ritmo cardíaco permitido, m, en latidos por minuto, puede ser aproximado por la ecuación: m= - 0.875 x + 190 donde x representa la edad del paciente de 1 a 99. Usando este modelo matemático determinar: a) El ritmo cardiaco máximo para una persona de 50 años. b) La edad de una persona cuyo máximo ritmo cardiaco sea de 160 latidos por minuto.
Engineering
1 answer:
Olin [163]3 years ago
3 0

Usando la ecuación de regresión que modela la frecuencia cardíaca máxima, podemos obtener el valor predicho de los problemas dados así:

La ecuación lineal que modela la frecuencia cardíaca máxima permitida en función de la edad del paciente está relacionada con la fórmula:

  • m = - 0,875x + 190

<em>x = edad del paciente; m = máx. frecuencia cardíaca permitida</em>

1.) <u>Frecuencia cardíaca máxima permitida para una persona de 50 años:</u>

Sustituye x = 50 en la ecuación:

m = -0,875 (50) + 190

m = 146,25

Por lo tanto, la frecuencia cardíaca máxima permitida es de aproximadamente 146 latidos / min.

2.) <u>Edad para una persona con frecuencia cardíaca máxima de 160 latidos / min</u>:

Sustituye m = 160 en la ecuación:

160 = -0,875x + 190

0,875x = 190 - 160

0,875x = 30

x = 30 / 0,875

x = 34,28

Por tanto, la edad de la persona sería de unos 34 años.

Más información: brainly.com/question/25395533

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Vesnalui [34]
<h3><u>CSMA/CD Protocol: </u></h3>

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But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

<u>Example: </u>

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12:00 AM A will see collisions

Pocket Size to detect the collision.

\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}

CSMA/CD is widely used in Ethernet.

<u>Efficiency of CSMA/CD:</u>

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We are given number of collisions (contentions slots)=4.

\text { Propagation day }=\frac{\text {distance}}{\text {speed}}

Distance = 1km = 1000m

\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}

7 0
3 years ago
A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a
nika2105 [10]

Answer:

Pressure = 115.6 psia

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Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, g_{c} = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

c_{p}= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence v_{a} ^{2} = 0

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} +  \frac{v_{1} ^{2} }{2c_{p} }

T_{1} = 20+460 = 480°R

T_{a}  =480+  \frac{(800)(800}{2(0.240)(25037}= 533.25°R

Pressure at the inlet of compressor at isentropic condition

P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}

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P_{2}= 8P_{a} = 8(14.45) = 115.6 psia

4 0
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kirill [66]

Solution :

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$V_1 = 7 \ m/s$

Operation time, $T_1$ = 3000 hours per year

$V_2 = 10 \ m/s$

Operation time, $T_2$ = 2000 hours per year

The density, ρ = $1.25 \ kg/m^3$

The wind blows steadily. So, the K.E. = $(0.5 \dot{m} V^2)$

                                                             $= \dot{m} \times 0.5 V^2$

The power generation is the time rate of the kinetic energy which can be calculated as follows:

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Regarding that $\dot m \propto V$. Then,

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Since, $\rho_a$ is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, $P_1= \text{const.} \times V_1^3$

            $P_1 = \text{const.} \times 343 \ W$

For the second site,

Power, $P_2 = \text{const.} \times V_2^3 \ W$

           $P_2 = \text{const.} \times 1000 \ W$

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3 years ago
What is the primary water source for a water cooled recovery unit's condensing coll?
nataly862011 [7]
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Naily [24]

Answer:

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Convert to stokes = 1.2727 stokes

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3 years ago
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