Answer:
Artefacts can influence our actions in several ways. They can be instruments, enabling and facilitating actions, where their presence affects the number and quality of the options for action available to us. They can also influence our actions in a morally more salient way, where their presence changes the likelihood that we will actually perform certain actions. Both kinds of influences are closely related, yet accounts of how they work have been developed largely independently, within different conceptual frameworks and for different purposes. In this paper I account for both kinds of influences within a single framework. Specifically, I develop a descriptive account of how the presence of artefacts affects what we actually do, which is based on a framework commonly used for normative investigations into how the presence of artefacts affects what we can do. This account describes the influence of artefacts on what we actually do in terms of the way facts about those artefacts alter our reasons for action. In developing this account, I will build on Dancy’s (2000a) account of practical reasoning. I will compare my account with two alternatives, those of Latour and Verbeek, and show how my account suggests a specification of their respective key concepts of prescription and invitation. Furthermore, I argue that my account helps us in analysing why the presence of artefacts sometimes fails to influence our actions, contrary to designer expectations or intentions.
When it comes to affecting human actions, it seems artefacts can play two roles. In their first role they can enable or facilitate human actions. Here, the presence of artefacts changes the number and quality of the options for action available to us.Footnote1 For example, their presence makes it possible for us to do things that we would not otherwise be able to do, and thereby adopt new goals, or helps us to do things we would otherwise be able to do, but in more time, with greater effort, etc
Explanation:
Technological artifacts are in general characterized narrowly as material objects made by (human) agents as means to achieve practical ends. ... Unintended by-products of making (e.g. sawdust) or of experiments (e.g. false positives in medical diagnostic tests) are not artifacts for Hilpinen.
Answer:
An oscilloscope display grid or scale is called a graticule.
Explanation:
it would be a bc its a sqare?well 3d is like you can say a cube 2d is like flat
Explanation:
Answer:
HEAT LOST
polycarbonate = 252 W
soda lime glass = 1680 W
aerogel = 16.8 W
COST associated with heat loss
polycarbonate = $ 262.08
soda lime glass = $ 1,747.2
aerogel = $ 17.472
The cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel
Explanation:
Given that;
surface area for each window = 0.4m * 0.4m = 0.16m^2
DeltaT = 90°C, L = 12mm = 0.012m
thermal conductivity of soda line can be gotten from tables in FUNDAMENTALS OF HEAT AND MASS TRANSFER
so at 300K
KsL = 1.4 W/mK
Kag = 0.014 W/mK
Kpc = 0.21 W/mK
Now HEAT LOSS
for polycarbonate;
Qpc = -KA dt/dx
NOTE ( heat flows from high temperature region to low temperature regions. so the second temperature would be smaller compared to the initial causing a negative in the change in temperature)
so Qag = (0.21 * 0.16 * 90) / 0.012
= 252 W
for soda lime glass;
Qsl = (1.4 * 0.16 * 90) / 0.012
= 1680 W
for aerogel
Qaq = (0.014 * 0.16 * 90) / 0.012
= 16.8 W
Now for COST associated with heat lost
for polycarbonate;
cost = Qpc * 130 * 8 * 1/1000
= 252 * 130 * 8 * 1/1000
= $ 262.08
for soda lime glass;
cost = 1680 * 130 * 8 * 1/1000
= $ 1,747.2
for aerogel
cost = 16.8 * 130 * 8 * 1/1000
= $ 17.472
Therefore the cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel
Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm
length of the photon emitted: 6.0 m
Explanation:
The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.
Recall the Heisenberg's uncertainty principle:-
∆t∆E ≈ h ( Planck's Constant)
The transition time ∆t corresponds to the energy that is ∆E
.
The corresponding uncertainty in the emitted frequency ∆v is:
∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1
To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate
λ = c/v
dλ/dv = -c/v² = -λ²/c
Therefore,
∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)
= 9.20 × 10⁻¹⁵ m or 9.20 fm
The length of the photon (<em>l)</em> is
l = (light velocity) × (emission duration)
= (3.0 × 10⁸ m/s)(20 × 10⁻⁹ s) = 6.0 m