Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
<u>T₂ = 95.56°C</u>
Answer:
The heat loss per unit length is 
Explanation:
From the question we are told that
The outer diameter of the pipe is 
The thickness is 
The temperature of water is 
The outside air temperature is 
The water side heat transfer coefficient is 
The heat transfer coefficient is 
The heat lost per unit length is mathematically represented as
![\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%5Cpi%20%28T%20-%20Ta%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_1%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_2%7D%7D)
Substituting values
![\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%2A%203.142%20%28363%20-%20263%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B300%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B20%7D%7D)
Answer:
the volume decreases at the rate of 500cm³ in 1 min
Explanation:
given
v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min
PV=C
vΔp + pΔv = 0
differentiate with respect to time
v(Δp/t) + p(Δv/t) = 0
(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0
40000 + 80kPa(Δv/t) = 0
Δv/t = -40000/80
= -500cm³/min
the volume decreases at the rate of 500cm³ in 1 min
Electron configurations:
Ge: [Ar] 3d10 4s2 4p2 => 6 electrons in the outer shell
Br: [Ar] 3d10 4s2 4p5 => 7 electrons in the outer shell
Kr: [Ar] 3d10 4s2 4p6 => 8 electrons in the outer shell
The electron affinity or propension to attract electrons is given by the electronic configuration. Remember that the most stable configuration is that were the last shell is full, i.e. it has 8 electrons.
The closer an atom is to reach the 8 electrons in the outer shell the bigger the electron affinity.
Of the three elements, Br needs only 1 electron to have 8 electrons in the outer shell, so it has the biggest electron affinity (the least negative).
Ge: needs 2 electrons to have 8 electrons in the outer shell, so it has a smaller (more negative) electron affinity than Br.
Kr, which is a noble gas, has 8 electrons and is not willing to attract more electrons at all, the it has the lowest (more negative) electron affinity of all three to the extension that really the ion is so unstable that it does not make sense to talk about a number for the electron affinity of this atom.
Answer:
A vacuum
Explanation:
Sound waves are examples of mechanical waves. Mechanical waves are waves which are transmitted through the vibrations of the particles in a medium.
For example, sound waves in air consist of oscillations of the air particles, which vibrate back and forth (longitudinal wave) along the direction of propagation of the wave itself.
Given this definition of mechanical wave, we see that such a wave cannot propagate if there is no medium, because there are no particles that would oscillate. Therefore, among the choices given, the following one:
a vacuum
represent the only situation in which a sound wave cannot propagate through: in fact, there are no particles in a vacuum, so the oscillations cannot occur. In all other cases, instead, sound waves can propagate.
