Physics, 23.03.2020 20:38 4300402428

Quiescent air is at 20°C (treat air as ideal gas). A plate of depth 50 cm and width 40 cm has a temperature of 40°C on the top surface. Calculate the heat dissipated from the top surface.

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Answer from: Orlandon220

Explanation:

The heat dissipated or loss at the top of the surface is given by

Qloss = -KA ∆T

Qloss = -KA (Ts - Ta)

Where

K is thermal conductivity of material

A is the cross-sectional area

L is the thickness or length of material

Ta is the temperature of air.

Ts is the temperature at the surface

Ts–Ta is the difference between skin temperature and air temperature.

Given that,

Temperature of air Ta=20°C

Plate dimensions are 50cm by 40cm

Converting to metre = 0.5m by 0.4m

Area = length × breadth

Area=0.5×0.4

Area=0.2m²

The temperature at surface

Ts=40°C

Let assume an aluminium plate then, k = 0.0263 W/m⋅K,

We don't need to convert the temperature to Kelvin, since we are going to use the change in temp.

Applying the formulas

Qloss = -KA (Ts - Ta)

Qloss = - 0.0263 × 0.2 × (40-20)

Qloss= -0.0263 × 0.2 × 20

Qloss= -0.1052J

The heat loss is 0.1052J

Answer from: Quest

increase masses, decrease separation

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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