Ask question

Ask question

All categories

3 years ago

12

Suppose you walk into a sauna that has an ambient temperature of 48.0°C. (a) Calculate the rate of heat transfer in watts to you

by radiation given your skin temperature is 37.0°C, the emissivity of skin is 0.98, and the surface area of your body is 1.30 m2.

1 answer:

GenaCL600 [577]3 years ago

4 0

Answer:99.84 W

Explanation:

Given

Ambient temperature $T_{\infty }=48^{\circ}C\approx 321\ K$

Temperature of skin $T=37^{\circ}C\approx 310\ K$

Skin emissivity $\epsilon =0.98$

Surface area $A=1.30\ m^2$

Rate of heat transfer is $Q=\epsilon A\sigma (T_0^4-T^4)$

$Q=0.98\times 1.30\times 5.67\times 10^{-8}((321)^4-(310)^4)$

$Q=99.84\ W$

You might be interested in

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.How muc

ahrayia [7]

Work is equal to the force applied times the displacement. Since you pull the wagon at constant speed this means that there is no acceleration on the wagon as it does not change speed. F=ma. Since a=0, F=0. Therefore no work has been done in this situation

3 0

3 years ago

Hi how do i run faster

Andrei [34K]

Answer:

Creatine is a great option , its not gonna help long term running ; but a great boost when you first start the run/jog.

Now to run faster you have to build great endurance and that is done through constant exercise running and stamina work outs such as sports , sprints calistetics etc.

Explanation:

Its not a simple 1 time thing and your suddenly , it takes drive and motivation !

Hope i was able to help you out ! Have a great day :)

7 0

2 years ago

PLEASE HELP WILL MARK AS BRAINLIEST!!!

marta [7]

The answer is a because tjhey do move at a molecular movemn

5 0

3 years ago

N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e

Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$

B) the speed of the sphere with mass 107.0 kg is

$v_2=\frac{m_1v_1}{m_2}$

$=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s$

C) the magnitude of the relative velocity with which one sphere is

$v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s$

D) the distance of the centre is proportional to the acceleration

$\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962$

Thus,

$x_1=3.962x_2$

and

$x_2=0.252x_1$

When the sphere make contact with eachother

Therefore,

$x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r$

And

$x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r$

The point of contact of the sphere is

$32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m$

3 0

3 years ago

How much heat is needed to raise the temperature of 8g of water by 20oC?

Elden [556K]

The following information are given in the question:
Mass, M = 8 g
Temperature, T = 20 degree Celsius
Specific heat of water [this value is a constant] C = 1 c/gc
Heat, Q = ?
The formula for calculating the amount of heat required is given below:
Q = MCT = 8 * 1 * 20 = 160
Therefore, Q = 160 cal.
<span />

7 0

3 years ago