Answer:
a
Solid Wire 
Stranded Wire 
b
Solid Wire 
Stranded Wire 
Explanation:
Considering the first question
From the question we are told that
The radius of the first wire is 
The radius of each strand is 
The current density in both wires is 
Considering the first wire
The cross-sectional area of the first wire is
= > 
= > 
Generally the current in the first wire is
=> 
=>
Considering the second wire wire
The cross-sectional area of the second wire is
=> 
=> 
Generally the current is
=> 
=>
Considering question two
From the question we are told that
Resistivity is 
The length of each wire is 
Generally the resistance of the first wire is mathematically represented as
=> 
=>
Generally the resistance of the first wire is mathematically represented as
=> 
=>
<span>The correct answer is C:Waves transfer energy, but not matter. A wave does not move matter in the direction of its propagation. It only transfers energy just like the ocean wave traveling many miles away with the water just moving up and down.</span>
Answer:
B. He should change the lengths of the vectors that point tangent to the circle so that each is the same length.
Explanation:
A uniform circular motion is a motion in a circle where the tangential speed of the object is constant.
In the motion map:
- The arrows pointing towards the centre of the circle represent the centripetal acceleration, and their length represent the magnitude of the acceleration
- The arrows pointing tangential to the circle represent the tangential speed, and their length represent the magnitude of the speed
In this motion map, we see that the length of the vectors pointing tangent to the circle is not constant: this means that the speed is not constant. In order to have a uniform circular motion, the speed must be constant, therefore the lengths of the vectors that point tangent to the circle must be the same.
Answer:
the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Explanation:
The torque is given by :
where ;
m = 0.160 A.m²
B = 0.0800 T
θ = 35°
So the magnitude of the torque N = mBsinθ
N = (0.160)(0.0800)(sin 35°)
N = 0.007341
N = 7.34×10⁻³ Nm
Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
b) The potential energy 
U = -mBcosθ
U = (- 0.160)(0.0800)(cos 45)
U = -0.010485
U = -1.0485 ×10⁻² J
Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Answer:
3A
Explanation:
Rtoal=R1+R2+R3=5+10+15=30
I=V/R 90/30
I=3
