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Elza [17]
3 years ago
10

When you cook a marshmallow on a metal poker tool over an open flame, energy is transferred. Identify the three different ways t

hermal energy is transferred in this situation. Be sure to name the specific places or phenomenon in this system where each of the types of thermal energy transfer is occurring and the direction the energy is being transferred (from what to what).

Physics
1 answer:
sergejj [24]3 years ago
4 0

Answer:

The three ways thermal energy is transferred are;

1) Conduction

2) Convection

3) Radiation

Explanation:

1) The conduction of the heat from the open flame to the marshmallow is through the direct contact of the flame with the marshmallow, such that the flame the region of the combustion reaction, that produces light and heat touches the marshmallow

2) The convection process is the transfer of heat from the rising heated combustion products, as well as the heated air that rises from the flame

3) The radiation heat transfer is the transfer of the heat from the fire to the marshmallows directly by the heat the moves in the form of electromagnetic waves at temperatures above 1000 K, without the need for a medium, such that the marshmallow can be heated by the heat coming from side of the flame.

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At t=0, a train approaching a station begins decelerating from a speed of 80 mi/hr according to the acceleration function a(t)=−
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Answer:

a)  Δx = 49.23 mi , b)  Δx = 5.77 mi

Explanation:

As we have an acceleration function we must use the definition of kinematics

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      v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt

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       v - v₀ = -1280 ∫ u⁻³ du / 8

       v -v₀ = -1280 / 8 (-u⁻²/2)

       v - v₀ = 80 (1+ 8t)⁻²

We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v

      v- 80 = 80 [(1 + 8t)⁻² - 1]

      v = 80 (1+ 8t)⁻²

We repeat the process for defining speed is

     v = dx / dt

    dx = vdt

    x-x₀ = 80 ∫ (1-8t)⁻² dt

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    x-x₀ = 80 (-1 / u)

    x-x₀ = -80 (1 / (1 + 8t))

We evaluate for t = 0 and x₀ and the upper point t and x

   x -x₀ = -80 [1 / (1 + 8t) - 1]

We already have the function of time displacement

a) let's calculate the position at the two points and be

t = 0 h

     x = x₀

t = 0.2 h

    x-x₀ = -80 [1 / (1 +8 02) -1]

    x-x₀ = 49.23

displacement is

  Δx = x (0.2) - x (0)

   Δx = 49.23 mi

b) in the interval t = 0.2 h at t = 0.4 h

t = 0.4h

     x- x₀ = -80 [1 / (1+ 8 0.4) -1]

     x-x₀ = 55 mi

    Δx = x (0.4) - x (0.2)

     Δx = 55 - 49.23

     Δx = 5.77 mi

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1.232\times 10^{-5}\ T.

Explanation:

<u>Given:</u>

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According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to \mu_o times the net current threading the loop.

\oint \vec B \cdot d\vec l=\mu_o I.

In case of a circular loop, the directions of magnetic field and the line element d\vec l, both are along the tangent of the loop at that point, therefore, \vec B\cdot d\vec l = B\ dl.

\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\

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Therefore,

B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_1 = 3.510\ m.

The magnetic field at the given point due to this wire is given by:

B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m,  concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_2 = 5.310\ m.

The magnetic field at the given point due to this wire is given by:

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The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at r_r=-3.510\ m is given by

B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.

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