How do the units of work and power compare?
1 answer:
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Answer:
d = 4.5079mm
Explanation:
from the question we are given that
= 0.082N/m
I₁= 30.6 A , I₂ = 60.4 A , and μ₀ = 4 π × 10⁻⁷ T · m / A
In order for the system to be in equilibrium, the repulsive magnetic force
per unit length on the top wire must equal the weight per unit length of the wire. Then we have
F/L = μ₀I₁I₂ / 2πd , making d as the subject of formular
we have that d = Lμ₀I₁I₂ / 2πF
= 4 π × 10⁻⁷ × 30.6 ×60.4 / 2π × 0.082
=45079.02× 10⁻⁷
in mm .. 45079.02× 10⁻⁷ × 10³
45079.02 ×10⁻⁴
d = 4.5079mm
Refer to the diagram shown below.
Define the (x,y) plane as the horizontal plane of the floor.
There was no momentum in the (x,y) plane before the plate hit the floor.
Let the velocity components in the (x) and (y) directions of the 100 g mass be Vx and Vy respectively, and that the resultant velocity, V, makes an angle θ below the negative x-axis as shown.
Because momentum is conserved, therefore
100*Vx + 320*2 = 0
100Vx = -640
Vx = -6.4 m/s
100Vy + 355*1.5 = 0
100Vy = -532.5
Vy = -5.325 m/s
V = √[(-6.4)² + (-5.325)²] = 8.33 m/s
θ = tan⁻¹ (-5.325/-6.4) = 39.8°
Answer:
The direction is 39.8° below the negative x-axis
The speed is 8.33 m/s
Define the (x,y) plane as the horizontal plane of the floor.
There was no momentum in the (x,y) plane before the plate hit the floor.
Let the velocity components in the (x) and (y) directions of the 100 g mass be Vx and Vy respectively, and that the resultant velocity, V, makes an angle θ below the negative x-axis as shown.
Because momentum is conserved, therefore
100*Vx + 320*2 = 0
100Vx = -640
Vx = -6.4 m/s
100Vy + 355*1.5 = 0
100Vy = -532.5
Vy = -5.325 m/s
V = √[(-6.4)² + (-5.325)²] = 8.33 m/s
θ = tan⁻¹ (-5.325/-6.4) = 39.8°
Answer:
The direction is 39.8° below the negative x-axis
The speed is 8.33 m/s