Answer:
d. 2.3 ohms (5.3 amperes)
Explanation:
The calculator's 1/x key makes it convenient to calculate parallel resistance.
Req = 1/(1/4 +1/8 +1/16) = 1/(7/16) = 16/7 ≈ 2.3 ohms
This corresponds to answer choice D.
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<em>Additional comment</em>
This problem statement does not tell the applied voltage. The answer choices suggest that it is 12 V. If so, the current is 12/(16/7) = 21/4 = 5.25 amperes.
Answer:
Sealing agent
Explanation:
Generally, when we have water leaks in almost any building or equipment, we use a sealant. However, this sealant could be of different types depending on the peculiarity of the leakage.
Thus, the correct answer is sealing agent.
Answer:
modulus of elasticity for the nonporous material is 340.74 GPa
Explanation:
given data
porosity = 303 GPa
modulus of elasticity = 6.0
solution
we get here modulus of elasticity for the nonporous material Eo that is
E = Eo (1 - 1.9P + 0.9P²) ...............1
put here value and we get Eo
303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )
solve it we get
Eo = 340.74 GPa
Answer:
Enthalpy of reaction (kJoules/mole)
Heat of formation of products (kJoules/mole)
Heat of reaction of reactants (kJoules/mole)
Explanation:
The general expression for calculating the overall enthalpy of reaction is given as following:
ΔH = ∑ΔH[producst] - ∑Δ[reactants]
Thus, the heat of reaction is given as the difference between the formation of the products and the formation of the reactants. The units are expressed as kJ/mol of reactants or products.
Thus, the three values are fundamental in the determination of the overall energy of the reaction from Hess' Law.
Answer:
The inductance of the inductor is 0.051H
Explanation:
From Ohm's law;
V = IR .................. 1
The inductor has its internal resistance referred to as the inductive reactance, X
, which is the resistance to the flow of current through the inductor.
From equation 1;
V = IX
X = 
................ 2
Given that; V = 240V, f = 50Hz, 
= 
, I = 15A, so that;
From equation 2,
X= 
= 16Ω
To determine the inductance of the inductor,
X = 2fL
L = 
= 
= 0.05091
The inductance of the inductor is 0.051H.
