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Sever21 [200]
2 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]2 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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Two points

Explanation:

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3 years ago
A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of
Naya [18.7K]

Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

Factor of safety FOS = 2.04

Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

(a) Find the solid length

Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in

Fs = k*Ys = k (Lo - Ls )

= 28(5 - 2.6) = 67.2 lbf Ans.

c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

= (4*10 + 2)/(4*10 - 3) = 1.135

Tau ts = Kb {(8FD)/(Πd^3)}

= 1.135 {(8*67.2*2)/(Π*2^3)}

= 48.56 * 10^6 psi

Let m = 0.187,

A = 147 kpsi.inm^3

Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

Ssy = 0.50 Sut

= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0
3 years ago
In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W
GrogVix [38]

Answer:

The shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

Explanation:

Given:

Temperature T = 1273.15 K

Initial Pressure P_{1} = 1.8 MPa

Final pressure P_{2} = 0.1 MPa

From the table superheated,

h_{i} = 4635 \frac{K J}{Kg} and  h_{f} = 2706.54 \frac{K J}{Kg}

Work done by shaft is,

 W = h_{f} - h_{i}

 W = 2706.54 - 4635

 W = -1928.46 \frac{kJ}{kg}

But here efficiency is 0.56,

So work generated per kg is,

Work = 0.56 \times(- 1928.46)

Work = -1.3 \frac{MJ}{kg}

Therefore, the shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

6 0
3 years ago
Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)
ArbitrLikvidat [17]

Answer:

\mathbf{P_x =25 \ watts}

\mathbf{x_{rmx} = 5 \ unit}

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) +  sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal  Asin (ωt), Power can be expressed as:

P= \dfrac{A^2}{2}

For the number of sinosoidial signals;

Power can be expressed as:

P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...

As such,

For x(t), Power  P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}

P_x = \dfrac{25}{2}+ \dfrac{25}{2}

P_x = \dfrac{50}{2}

\mathbf{P_x =25 \ watts}

For the number of sinosoidial signals;

RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...

For x(t), the RMS value is as follows:

x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }

x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }

x_{rmx }=\sqrt{(\dfrac{50}{2} )}

x_{rmx} =\sqrt{25}

\mathbf{x_{rmx} = 5 \ unit}

8 0
3 years ago
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Effectus [21]
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3 years ago
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