The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.
- The force per unit length between two parallel thin current-carrying
and 
wires at distance ' r ' is given by 
....(1) .
- If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.
A schematic of the information provided in the question can be seen in the image attached below.
From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3
Using equation (1) , we get
I₃ = 2.4 A and the current is pointing in the downward direction
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Answer:
Ang answer units of J heat of fusion is 3.33 x 105
Answer:
B. The maximum angle decreases
Explanation:
If θ be the maximum angle of a slope that allows a crate placed on it to remain at rest , following condition exists .
tanθ = μ , θ is called angle of repose . μ is coefficient of static friction .
So the tan of angle of repose θ is proportional to coefficient of static friction.
If coefficient of static friction is less than .7 , naturally angle of repose will also become less ,ie, it at lower angle of inclination , the object will start slipping .
volume of balloon
= 4/3 T R3
= 4/3 x 3.14 x 6.953
= 1405.47 m3
uplift force
= volume of balloon x density of air x 9.8
= = 1405.47 x 1.29 x 9.8
= 1813.05 x 9.8 N
weight of helium gas
= volume of balloon x density of helium x
9.8
= 1405.47 x .179 x 9.8
= 251.58 x 9.8 N
Weight of other mass = 930 x 9.8 N Total weight acting downwards
= 251.58 x 9.8 +930 x 9.8
= 1181.58 x 9.8 N
If W be extra weight the uplift can balance
1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8
1181.58+W=1813.05
W= 631.47 kg
