If the primary wire's power is 10 A and one branch's power is 4 A, another branch's power will be 6A.
According to Kirchhoff's current law (KCL), the total current flowing through a parallel route circuit's junction equals the total current flowing away from it.
Provided that one of the two branches through which power exits the intersection has a flow of 4A, and also that the junction's overall flow entering it is 10A, the entire current going the junction should be 10A.
Consequently, the second wire's power may be expressed as;
I = I1+ I2 [ where I= total current (10A);
I1= current in one branch (4A) &
I2= current in another branch]
⇒I2 = I - I1
⇒I2 = 10A - 4A
⇒I2 = 6A
Therefore, it can be concluded that when the primary wire bears 10A power having 4A in one of its branches, another branch carries 6A power.
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The first thing you should know for this case is that density is defined as the quotient between mass and volume.
d = m / v
We have two states:
State 1:
d1 = 1.20 g / l
v1 = 1.02 × 106 l
State 2:
v2 = 1.09 × 106 l
Since the mass remains constant, then:
m = d1 * v1
Then, the density in state two will be:
d2 = m / v2
Substituting the value of the mass we have:
d2 = (d1 * v1) / v2
Substituting the values:
d2 = ((1.20) * (1.02 * 10 ^ 6)) / (1.09 * 10 ^ 6) = 1.12 g / l
answer:
The density of the heated air in the balloon is 1.12 g / l
Answer:
-2.66 kJ
Explanation:
At constant pressure, work done is:
W = -P ΔV
W = -(95.000 Pa) (43 m² − 15 m²)
W = -2660 J
W = -2.66 kJ
