Conducting plates will be 3.75 m apart if an electric field strength of 4. 4 kV/m between them, if their potential difference is 15 kV
The potential difference, also referred to as voltage difference between two given points is the work in joules required to move one coulomb of charge from one point to the other. The SI unit of voltage is the volt. Volt Formula.
In a simple parallel-plate capacitor, a voltage applied between two conductive plates creates a uniform electric field between those plates. The electric field strength in a capacitor is directly proportional to the voltage applied and inversely proportional to the distance between the plates.
E = V/d
E = Electric field strength
d = distance between the plates
V = potential difference
Electric field strength = 4 kV/m
Potential difference = 15 kV
d = V / E = 15 kV / 4 kV/m
= 3.75 m
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Answer:
e. All of these statements are false.
Explanation:
As we know that heat transfer take place from high temperature to low temperature.
It is possible to convert all work into heat but it is not possible to convert all heat in to work some heat will be reject to the surrounding.
The first law of thermodynamics is the energy conservation law.
Second law of thermodynamics states that it is impossible to construct a device which convert all energy into work without rejecting the heat to the surrounding.
By using heat pump ,heat can transfer from cooler body to the hotter body.
Therefore all the answer is False.
Answer:
T = 0.0088 m²/s
Explanation:
given,
initial piezometric elevation = 12.5 m
thickness of aquifer = 14 m
discharge = 28.24 L/s = 0.02824 m³/s
we know
k = 0.629 mm/sec
Transmissibilty
T = k × H
T = 0.629 × 14 × 10⁻³
T = 0.0088 m²/s
Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²
