Answer:
Qcd=0.01507rad
QT= 0.10509rad
Explanation:
The full details of the procedure and answer is attached.
Answer:
In the simplest terms, bioremediation is a waste management process using live organisms to neutralize or remove harmful pollutants from contaminated areas. Bioremediation is an environmental science that amplifies natural biological actions to remedy or remediate polluted groundwater and contaminated soil.
Explanation:
Answer:
The coefficient of thermal expansion tells us how much a material can expand due to heat.
Explanation:
Thermal expansion occurs when a material is subjected to heat and changes it's shape, area and volume as a result of that heat. How much that material changes is dependent on it's coefficient of thermal expansion.
Different materials have different coefficients of thermal expansion (i.e. It is a material property and differs from one material to the next). It is important to understand how materials behave when heated, especially for engineering applications when a change in dimension might pose a problem or risk (eg. building large structures).
Use protective gear. Use insulated tools, Wear flame resistant clothing, safety glasses, and insulation gloves, Remove watches or other jewelry, Stand on an insulation mat. 03. Never connect the insulation tester to energized conductors or energized equipment and always follow the manufacturer's recommendations. When installing new electrical machinery or equipment, testing insulation resistance is important for two reasons. First, it ensures that the insulation is in adequate condition to begin operation. ... The test is accomplished by applying DC voltage through the de-energized circuit using an insulation tester. Insulation resistance should be approximately one megohm for each 1,000 volts of operating voltage, with a minimum value of one megohm. For example, a motor rated at 2,400 volts should have a minimum insulation resistance of 2.4 megohms.
Answer:
The required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
.
Explanation:
We are given current density of 459 A/cm² and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.
Recall that current density is given by
j = I/A
where I is the current flowing through the wire and A is the area of the wire
A = πr²
but r = d/2 so
A = π(d/2)²
A = πd²/4
so the equation of current density becomes
j = I/πd²/4
j = 4I/πd²
Re-arrange the equation for d
d² = 4I/jπ
d = √4I/jπ
d = √(4*0.53)/(459π)
d = 0.0383 cm
Therefore, the required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
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