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: 361° C
Explanation:
Given
Initial pressure of the gas, P1 = 294 kPa
Final pressure of the gas, P2 = 500 kPa
Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K
Final temperature of the gas, T2 = ?
Let us assume that the gas is an ideal gas, then we use the equation below to solve
T2/T1 = P2/P1
T2 = T1 * (P2/P1)
T2 = (100 + 273) * (500 / 294)
T2 = 373 * (500 / 294)
T2 = 373 * 1.7
T2 = 634 K
T2 = 634 K - 273 K = 361° C
Answer:
Our drinking water comes from lakes, rivers and groundwater. For most Americans, the water then flows from intake points to a treatment plant, a storage tank, and then to our houses through various pipe systems. A typical water treatment process.
Explanation:
Explanation:
Value of the cross-sectional area is as follows.
A = 
= 3.45 
The given data is as follows.
Allowable stress = 14,500 psi
Shear stress = 7100 psi
Now, we will calculate maximum load from allowable stress as follows.
= 
= 50025 lb
Now, maximum load from shear stress is as follows.
= 
= 48990 lb
Hence, 
will be calculated as follows.
= 48990 lb
Thus, we can conclude that the maximum permissible load is 48990 lb.
