Complete question:
Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.
Answer:
The bottom current is 12.8 A to the right.
Explanation:
Given;
length of the wires, L = 3.0 m
current in the top wire, I₁ = 12.5 A
repulsive force between the two wires, F = 2.4 x 10⁻⁴ N
distance between the two wires, r = 40 cm = 0.4 m
The repulsive force between the two wires is given by;
Where;
I₂ is the bottom current
The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.
Therefore, the bottom current is 12.8 A to the right.
Answer: 2 cm
Explanation:
Given , for a converging lens
Focal length : 
Height of object : 
Object distabce from lens : 
Using lens formula: 
, we get
, where v = image distance from the lens.
On solving aboive equation , we get
Formula of Magnification : 
, where h' is the height of image.
Put value of u, v and h in it , we get
Hence, the height of the image is 2 cm.
Answer:
force-strength,power or energy as an attribute of motion, movement or action. Example: Frictional force.
Volume of gold in the phone = 10 cm^3
= 0.<span>00001 m^3 </span>
Density of gold = 19300 kg/m^3
1 kg mass = 2.2 pounds
Mass of 10 cm^3 of gold = 0<span>.00001 m^3 * (19300 kg/m^3)
= 0.193 kg
So
0.193 kg = 0.193 * 2.2 pounds
= 0.43 pounds
I think there is something wrong with the options given in the question.</span>
The final velocity is 5.87 m/s
<u>Explanation:</u>
Given-
mass, 
= 72 kg
speed, 
= 5.8 m/s
,
= 45 kg
,
= 12 m/s
Θ = 60°
Final velocity, v = ?
Applying the conservation of momentum:
X + X = ( + ) v
72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v
v = 417.6 + 540 X 
v = 417.6 + 
v = 5.87 m/s
The final velocity is 5.87 m/s
