That is too hard but u got that cuz i believe in you!!!
Answer:
61.5 °C
Explanation:
The resistance of the relay coil at 15 °C is R = V/I = 6/0.12 = 50 ohms. In order for the coil current to remain above 0.10 A, the resistance must remain below R = 6/0.10 = 60 ohms.
At some temperature difference ΔT from 15 °C, the resistance of the coil will be ...
R = R0(1 +α·ΔT)
where R0 is the resistance at 15 °C, α is the temperature coefficient of resistance, and ΔT is the temperature change. We want to solve this for ΔT:
R/R0 = 1 +α·ΔT
(R/R0 -1)/α = ΔT = (60/50 -1)/0.0043 ≈ 46.5 . . . . °C
The relay may fail to operate at temperatures above (15 +46.5) °C = 61.5 °C.
Answer:
An airfoil creates lift by exerting a downward force on the air as it flows past
Answer:
(a)This portion of the problem asks that we calculate, for a Pb-Mg alloy, the mass of lead in 7.5 kg of thesolidphase at 300C just below the solubility limit.From Figure 9.20, the solubility limit for thephase at
Explanation:
Answer and Explanation:
Core self-evaluations (CSEs) stands for a wide personality trait that comprises of 4 positive individual traits, namely:
(1) self-efficacy
(2) self esteem
(3) locus of control
(4) emotional stability.
Baiscally, when people have a positive evaluation about themselves, or quality core self-evaluation, they believe that they are worthy and fit for a task. They trust their capability and effectiveness. This leads to some implications in their managers duties and careers, which could either be positive or negative.
A group leader can use CSEs to create a more effective unit by implementing the ten items points of Generalized Self Efficacy Tool to test the self efficiency of individual personnels in that unit.