What is meant by the term law of conservation?

Physics

2 answers:

GalinKa [24]2 years ago

4 0

Answer:

Explanation:

any law stating that some quantity or property remains constant during and after an interaction or process, as conservation of charge or conservation of linear momentum.

padilas [110]2 years ago

4 0

<h2><em><u>Answer:</u></em></h2><h2><em><u>The law of conservation of matter or principle of matter conservation states that the mass of an object or collection of objects never changes over time, no matter how the constituent parts rearrange themselves. The mass can neither be created nor destroyed.</u></em></h2><h2><em><u>Explanation:</u></em></h2>

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The temperature of the cosmic background radiation is measured to be 2.7 k. What is the wavelength of the peak in the spectral d

KATRIN_1 [288]

Answer:

$1.07\cdot 10^{-3} m$

Explanation:

The peak wavelength of the spectral distribution can be found by using Wien's displacement law:

$\lambda=\frac{b}{T}$

where

$b=2.898\cdot 10^{-3} m\cdot K$ is Wien's displacement constant

T is the absolute temperature

For the cosmic background radiation, the temperature is

T = 2.7 K

So, the corresponding peak wavelength is

$\lambda=\frac{2.898\cdot 10^{-3} m\cdot K}{2.7 K}=1.07\cdot 10^{-3} m$

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What would it be like to walk on Venus?

maria [59]

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3 years ago

A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$