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2 years ago

Is our body in thermal equilibrium with the environment ?

Physics

1 answer:

LenKa [72]2 years ago

4 0

Answer:

The human body runs at a constant 37 ºC but the air around you at room temperature is about 20-25 ºC which means heat is constantly leaving your body to warm your surroundings and maintain thermal equilibrium. You don't lose much energy doing this however as air heats reasonably quickly

Explanation:

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3 years ago

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2 years ago

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A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

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2 years ago

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2 years ago

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satela [25.4K]

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2 years ago

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Is our body in thermal equilibrium with the environment ?​

Is our body in thermal equilibrium with the environment ?