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11Alexandr11 [23.1K]
3 years ago
6

The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A 157 lb (71.0 kg ) perso

n of height 5.91 ft(1.80 m ) would have a body surface area of approximately 1.90 m2 .
Reqiuired:
a. What is the net amount of heat this person could radiate per second into a room at 19.0 ∘C (about 66.2∘F) if his skin's surface temperature is 31.0 ∘C? (At such temperatures, nearly all the heat is infrared radiation, for which the body's emissivity is 1.00, regardless of the amount of pigment.)

b. Normally, 80.0 % of the energy produced by metabolism goes into heat, while the rest goes into things like pumping blood and repairing cells. Also normally, a person at rest can get rid of this excess heat just through radiation. Use your answer to part A to find this person's basal metabolic rate(BMR).
Physics
1 answer:
GalinKa [24]3 years ago
6 0

Answer:

A. Net amount of heat radiated = 109.2W

B. Person's basal energy = 136.5

Explanation:

Part A:

Area of person, A = 1.90 m^2

Temperature of person , T = 31 C

T = 304 K

Temperature of surroundings , To = 19 C

To = 282 K

Now, net amount of heat radiated = e*A*sigma *(T^4 - To^4)

Net amount of heat radiated = 1 * 1.8 * 5.6703 *10^-8 *(304^4 - 294^4)

Net amount of heat radiated = 109.2 W

The net amount of heat radiated is 109.2 W

Part B:

Person's basal energy = net amount of heat radiated /(0.80)

Person's basal energy = 109.2/0.80

Person's basal energy = 136.5 W

Person's basal energy is 136.5 W

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Answer:

The distance is D  =  2.6 \ m

Explanation:

From the question we are told that

    The wavelength of the light is  \lambda  =  476.1 \ nm  =  476.1 *10^{-9} \ m

      The  distance between the slit is  d =  0.29 \  mm  =  0.29 *10^{-3} \ m

       The  between the first and second dark fringes is  y =  4.2 \ mm  =  4.2 *10^{-3} \ m

Generally  fringe width is mathematically represented as

       y  =  \frac{\lambda * D }{d}

Where D is the distance of the slit to the screen

   Hence

        D  =  \frac{y *  d}{\lambda }

substituting values

       D  =  \frac{ 4.2 *10^{-3} *   0.29 *10^{-3}}{ 476.1 *10^{-9} }

        D  =  2.6 \ m

7 0
3 years ago
A mass on a spring with k=88.7 N/m oscillates 15 times in 9.24s. what is the objects mass? unit=kg?
sweet [91]

The mass on the spring is 0.86 kg

Explanation:

The period of a mass-spring system is given by the equation

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant

In this problem, we have:

k = 88.7 N/m is the spring constant

The system makes 15 oscillations in 9.24 s: therefore, the period of the system is

T=\frac{9.24}{15}=0.62 s

Now we can re-arrange the first equation  to solve for the mass:

m=k(\frac{T}{2\pi})^2=(88.7)(\frac{0.62}{2\pi})^2=0.86 kg

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brainly.com/question/5438962

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A soccer ball kicked with a force of 12.5 N accelerates at 6.2 m/s’ to the right. What is the mass of the ball? Answer in units
stepladder [879]

Answer:

m = 2.01[kg]

Explanation:

This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.

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F = force = 12.5 [N]

m = mass [kg]

a = acceleration = 6.2 [m/s²]

12.5=m*6.2\\m = 2.01[kg]

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A 10 kg box is 1.3 m above the ground. How much potential energy does it have? (g on Earth of 9.8 m/s?
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