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In-s [12.5K]
3 years ago
6

The total surface area of the human body is 1.20 m2 and the surface temperature is 30∘C=303∘K. If the surroundings are at a temp

erature of 4.0 ∘C , what is the net rate of heat loss from the body by radiation? The emissivity of the body is very close to unity, irrespective of skin pigmentation. Express your answer using two significant figures.
Physics
1 answer:
zalisa [80]3 years ago
8 0

To solve this problem we must consider the expressions of Stefan Boltzmann's law for which the rate of change of the radiation of energy H from a surface must be

H = Ae\sigma T^4

Where

A = Surface area

e = Emissivity that characterizes the emitting properties of the surface

\sigma = Universal constant called the Stefan-Boltzmann constant (5.67*10^{-8}m^{-2}K^{-4})

T = Absolute temperature

The total heat loss would be then

Q = H_2 -H_1

Q =Ae\sigma T_2^4-Ae\sigma T_1^4

Q = Ae\sigma (T_2^4-T_1^4)

Q = (1.2)(1)(5.67*10^{-8})(303^4-280^4)

Q = 155.29J

Therefore the net rate of heat loss from the body by radiation is 155.29J

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3 years ago
Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and
kow [346]

A) Acceleration: a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

A)

Acceleration is defined as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first  segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

a_1 = \frac{11-8}{4}=0.75 m/s^2 (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: a_2 = 0

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

a=\frac{0-11}{7}=-1.57 m/s^2

And the negative sign means the acceleration is south, opposite to the direction of motion.

B)

In a uniformly accelerated motion, the displacement can be calculated as:

s=ut+\frac{1}{2}at^2

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

u = 0\\a = 0.75 m/s^2\\t=4 s

So the displacement is

s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m

- For the second segment, we have

u = 11 m/s\\a = 0\\t=15 s

So the displacement is

s_2 = (11)(15)+0=165 m

- For the third segment, we have

u = 11\\a = -1.57 m/s^2\\t=7 s

So the displacement is

s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

C)

The average velocity is given by:

v=\frac{d}{t}

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

v=\frac{209.5}{26}=8.06 m/s (north)

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7 0
3 years ago
A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera
yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

\Delta V_{max} = 240 V

frequency, f = 50Hz

I_{max} = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, \omega =  2\pi f = 2\pi \times50 = 100\pi

a). Inductive reactance,  X_{L} is given by:

\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega

X_{L} = 47.12\Omega 

b). The capacitive reactance,  X_{C} is given by:

\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega

X_{C} = 0.636\Omega

c). Impedance, Z = \frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega

Z = 2400\Omega

d). Resistance, R is given by:

Z = \sqrt {R^{2} + (X_{L} - X_{C})}

2400^{2} = R^{2} + (47.12 - 0.636)^{2}

R = \sqrt {5757839.238}

R = 2399.5\Omega

e). Phase angle between current and the generator voltage is given by:

tan\phi = \frac{X_{L} - X_{C}}{R}

\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})

\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}

\phi = 1.11^{\circ}

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3 years ago
Samples of different materials, A and B, have the same mass, but the sample
Effectus [21]

Answer:

B. The particles that make up material B have more mass than the

particles that make up material A.

Explanation:

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3 years ago
Which of the following tools is used to observe objects that are very far away? A. a anemometer B. a hand lens C. a microscope D
inn [45]

Answer:

D. a telescope

Explanation:

3 0
3 years ago
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