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Yanka [14]
2 years ago
7

Compare the maximum rate of heat transfer to the basal metabolic rate by converting a bmr of 88 kcal/hr into watts. what is the

value? from this comparison you should see why the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat [12].
Physics
1 answer:
elena-14-01-66 [18.8K]2 years ago
8 0

Explanation :

It is given that,

BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :

We know that, 1 kilocalorie = 4184 joules

So, 1\ kcal/h=\dfrac{1\times 4184\ J}{3600\ sec}

1\ kcal/h=1.16\ J/sec

J/sec is nothing but watts.

So, 1\ kcal/h=1.16\ watts

and 88\ kcal/h=88\times 1.16\ watts = 102.08\ watts

So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.

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An object weighs 980N on the earth’s surface (i) What is its mass? (ii) If the same object weighs 360N on another planet, find t
AnnZ [28]
Weight = mass*gravity. Hence mass = 980/9.8 = 100kg. Gravity of planet 2 = weight/mass = 3.6 m/s^2
4 0
2 years ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
2 years ago
A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr
Harlamova29_29 [7]

Answer:

a) m = 3 for λ1 and m= 2 for λ2 will overlap since they have the same values

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The detailed steps and appropriate formula is as shown in the attachment.

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A capacitor with C = 6.00 μF is fully charged by connecting it to a battery that has emf 50.0 V. The capacitor is disconnected f
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Answer:

1.99×10^-4coulombs

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The charge (Q) across the resistor the directly proportional to the voltage (V) where capacitance of the capacitor(C) is the proportionality constant. Mathematically, Q = CV

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The charge Q on the resistor will be;

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Were C = 6.00 μF, V = 33.3

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Q= 1.99×10^-4Coulombs

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Answer:

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