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3 years ago

How could the campfire warm things by conduction and radiation?

Physics

2 answers:

Pie3 years ago

7 0

Answer:

Explanation:

Radiation is what occurs when energy is transmitted or emission of energy through the space or in the form of waves. This includes: electromagnetic radiation, such as radio waves, microwaves, infrared, visible light, ultraviolet, x-rays, and gamma radiation

Conduction is a a more so physical contact situation but if connection to the ground is made then the heat can transfer into the ground and proceed to spread into a wider vicinity warming the general area around it in a more physical manner.

liubo4ka [24]3 years ago

5 0

Answer:

The heat can tranfer both directly and through electromagnetic waves.

Explanation:

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3 years ago

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4 years ago

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3 years ago

Exit which rock is not sedimentary? a. shale b. schist c. limestone d. sandstone

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4 years ago

A spring with a 6-kg mass and a damping constant 7 can be held stretched 0.5 meters beyond its natural length by a force of 1.5

Scilla [17]

Answer:

The value for $c^2 - 4mk$ is : $\mathbf{-23 \ m^2kg^2/sec^2}$

The position of the mass (m) after t seconds is:

$\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}$

Explanation:

The spring constant is :

$F = kx \\ \\ k = \frac{F}{x} \\ \\ k = \frac{1.5}{0.5} \\ \\ k = 3 \ N/m$

The value for $c^2 - 4mk$ is :

= $7^2 - 4(6)(3)$

= $49 - 27$

= $\mathbf{-23 \ m^2kg^2/sec^2}$

The differential equation for this system is :

$m \frac{d^2x}{dt^2}+c \frac{dx}{dt}+kx = 0$

$6 \frac{d^2x}{dt^2}+7\frac{dx}{dt}+3x = 0$

and the auxiliary equation for this differential equation is :

$6m ^2+ 6m + 3 = 0$

using the quadratic formula :

$\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

$\frac{-6 \pm \sqrt{6^2 - 4(6)(3)} }{2(6)}$

= $\frac{-6 \pm \sqrt{-36} }{12}$

= $-\frac{1}{2} \pm \frac{\sqrt{36} }{12}i$

The general solution is :

$x(t) = e^{-\frac{1}{2}t}}(c_1 cos {\frac{\sqrt{36} }{12}} t + c_2 sin \frac{\sqrt{36} }{12} t})$

Initial conditions : $x(0) = 1 \ m , x' (0) = 0\\$

$x(0) = ( c_1 cos 0 + c_2 sin 0)$

$1 = c_1$

$x't = e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin \frac{\sqrt{36} }{12}t + c_2 \frac{\sqrt{36} }{12} cos \frac{\sqrt{36} }{12}t) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_1cos \frac{\sqrt{36} }{12} t +c_2 sin \frac{\sqrt{36} }{12} t)$

$x'(0)= e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin0 + c_2 \frac{\sqrt{36} }{12} cos 0) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_10+c_2 sin0)$