A closed circle means the number is included and an open circle means its not.
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:
1. 18.5m/s
2. 17.5 m
3. 0 at its highest point
4. Direction is downwards
Explanation:
1. This egg is thrown vertically from a height
Yo = 0. This egg then falls to the point y = -30.0 at t = 5seconds
Y-Yo = V0t - 1/2gt²
-30-0 = V0(5)-1/2(9.8)(5²)
-30 = 5v0 - 4.9x25
-30 = 5V0 - 122.5
-30+122.4 = 5v0
V0 = 92.5/5
= 18.5m/s
<em><u>this </u></em><em><u>is </u></em><em><u>the </u></em><em><u>initial</u></em><em><u> </u></em><em><u>speed</u></em><em><u> of</u></em><em><u> the</u></em><em><u> </u></em><em><u>egg</u></em>
2. When the egg is at a maximum height it would have a velocity equal to 0
V² = V0² - 2*g*y
V = 0, V0 = 18.5, g = 9.8
0 = 18.5²-2x9.8*y
342.25-19.6y = 0
342.25 = 19.6y
Divide through by 19.6
Y = 342.25/19.6
Y = 17.5m
<em><u>this value is how high it rises above starting point</u></em>
3.
The magnitude of velocity is = 0 at its highest point
4.
This egg falls under gravity. Therefore the acceleration due to gravity has a constant magnitude and direction. Magnitude = 9.8m/s and it's direction is downwards.
5. Please check attachment for graph
Answer:
4°
Explanation:
Given,
chord in the airstream, c = 2 m
speed, v = 50 m/s
lift per unit span = 1353 N/m
density of air, ρ = 1.225 kg/m³
Lift,



From the plot of c_L and angle of attack
Angle of attack is equal to 4°
Explanation:
o has Celsius of indirect measurment