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3 years ago

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Expression for calories burned during workout

1 answer:

VMariaS [17]3 years ago

8 0

Women: Calories -((Age x 0.074) - (Weight x 0.05741) + (Heart Rate x 0.4472) – 20.4022) Time / 4.184 Men: Calories = ((Age x 0.2017) + (Weight 0.09036) + (Heart Rate x 0.6309) - 55,0969) x Time / 4.184

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The technique of smoothing out joint compound on either side of a joint is known as which of the following

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a is the answer to your questions

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3 years ago

Why did you choose agricultural and biosystem as a course? Help me guys

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To be able to develop as a human and give back to the community. To ensure that people have safe food and water to drink, clean fuel and energy sources, and in general a safe enviornment to live in.

Explanation:

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3 years ago

a centrifugal pump delivers water at the rate of 50kg/is . the inlet and outlet pressure are 2 bar and 6.2 bar respectively. the

Monica [59]

Answer:

21.6 kw

Explanation:

Given data:

m = 50 kg/s

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outlet pressure(p2) = 6.2 bar

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d2 = 100 mm = 0.1 m

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next we have to determine the Q value

Q = V*A

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next we have to calculate the various V's

V1 = Q/A1 = $\frac{0.05*4}{\pi *0.2^2}$ = 1.59 m/s

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Determine the capacity of the electric motor

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3 years ago

You are to assess the biomechanics of a male’s arm using his bicep to hold a 20 kg object in his hand. The upper arm is perpendi

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4 years ago

A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an

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Answer:

The right solution is "28.45%".

Explanation:

The given values are:

$P_4=50\ kPa$

$h_4=0.7(2304.7)+340.5$

$=1953.83 \ KJ/Kg$

and,

$P_3=15 \ mPa$

$h_3=hg$

$=2610.8 \ KJ/Kg$

$s_3=sg$

$=5.3108 \ KJ/Kgh$

At 45,

⇒ $x_{45} = \frac{5.3108-1.0912}{6.5019}$

$=0.66$

At $P_4=50 \ Kpa$ ,

$h_f=340.54$

or,

$V_f=0.001030 \ m^3/Kg$

then,

⇒ $h_2=340.54+0.001030(15\times 10^{3}-50)$

$=355.94 \ kJ/kg$

hence,

The isentropic efficiency of turbine will be:

⇒ $n_T=\frac{h_3-h_4}{h_3-h_{45}}$

$=\frac{2610.8-1953.83}{2610.8-1836.26}$

$=84.818$ (%)

The thermal efficiency of cycle will be:

⇒ $n_C=\frac{W_T-W_P}{2_{in}}$

$=\frac{(2610.8-1953-83)-(355.93-340.54)}{2610.8-355.93}$

$=28.45$ (%)

4 0

3 years ago