C.) Both male and female
Hope this helped!
The magnitude of the magnetic force per unit length on the top wire is
2×10⁻⁵ N/m
<h3>How can we calculate the magnitude of the magnetic force per unit length on the top wire ?</h3>
To calculate the magnitude of the magnetic force per unit length on the top wire, we are using the formula
F=
Here we are given,
= magnetic permeability
= 4×10⁻⁷ H m⁻¹
If= 12 A
d= distance from each wire to point.
=0.12m
Now we put the known values in the above equation, we get
F=
Or, F =
Or, F= 2×10⁻⁵ N/m.
From the above calculation, we can conclude that the magnitude of the magnetic force per unit length on the top wire is 2×10⁻⁵ N/m.
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Answer: The train traveled 210 miles in 6 seconds of time duration.
Explanation:
The initial velocity of the train,u = 50 mile/s
Acceleration of the train ,a =
Duration of time ,t = 6 seconds
Using second equation of motion :
The train traveled 210 miles in 6 seconds of time duration.
Your answer would be i think repetition
Answer:
8N and 32N
Explanation:
Given that a light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center.
To calculate the forces that are exerted by the sawhorses on the board, we must consider the equilibrium of forces acting on the board.
Let the two upward forces produce by the saw horses be P1 and P2
Assuming that the weight is negligible
Sum of the upward forces = sum of the downward forces.
P1 + P2 = 40 ....... (1)
Also, the sum of the clockwise moment = sum of the anticlockwise moments.
Let's assume that the board is uniform. The weight will act at the centre.
Taking moment at the centre:
P1 × 5 + 40 × 2 = 0
P1 = 40 / 5
P1 = 8N
Substitute P1 into equation 1
8 + P2 = 40
P2 = 40 - 8
P2 = 32N