The magnitude of the current in wire 3 is (I₃)= 0.33A
<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>
To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,
I₁ + I₂ + I₃ = 0
Where we are given,
I₁ = current in wire 1
=0.40 A.
I₂ = current in wire 2
= -0.73 A.
We have to calculate the magnitude of the current in wire 3, I₃
Now we put the known values in above equation, we get,
I₁ + I₂ + I₃ = 0
Or, I₃ = -.(I₁ + I₂)
Or, I₃ = -.(0.40 - 0.73)
Or, I₃ = 0.33 A
From the above calculation, we can conclude that the current in wire 3 is I₃ = 0.33 A
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Answer:
69.69 g
Explanation:
Evaporation of water will take out latent heat of vaporization. Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg
Heat taken up by evaporating water
= 2260000 x m J
Heat lost by body
= mass x specific heat of body x drop in temperature
60 x 3500 x .750 ( specific heat of human body is 3.5 kJ/kg.k)
= 157500 J
Heat loss = heat gain
2260000 m= 157500
m = .06969 kg
= 69.69 g
We know, F = 1/4πε * q₁q₂ / r²
Here, q₁ = 5 * 10⁻⁶ C
q₂ = 2 * 10⁻⁶ C
r = 3 * 10⁻² m west
Substitute their values,
F = (9 * 10⁹) (5 * 10⁻⁶) (2 * 10⁻⁶) / (3 * 10⁻²)²
F = 100 N [ East of positive charge ]
Hope this helps!
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Born: April 7, 1972 (age 44), Chichester, United Kingdom
Space missions: Expedition 46, Expedition 47, Soyuz TMA-19M
Nationality: British
Answer:
The potential difference between the student and the tap causes electrons to transfer to the tap which is earthed.
Hope this helps my sis and others
-Amelia
