Answer:
0.446 mm
0.066 V/m
Explanation:
Given
We are given the length of the copper cable L = 3.30 km and the potential difference is V = 220 V
Solution
(a) We want to find the diameter d of the cable when the dissipated power is P = 50W. The power consumed by the cable depends on its resistance R and it is given by equation in the form
P= V^2/R (1)
Where V is the voltage in the cable. Now let us solve equation (1) for R and plug our values for V and P into equation (1) to get R
R = V^2/P = (220)^2/(50) = 968Ω
Now we can determine the diameter of the copper wire. The resistance R of the wires depends on the area of the wire, resistivity and the length of the cable. Where equation gives us the relationship between these variables in the form
R = pL/π*r^2 (solve for r)
r = √pL/πR (2)
Now we can plug our values for Rep and L into equation (2) to get the radius of the cable where p for copper equals 1.72 x 10-8 Ω m
r =√pL/πR
= √1.72 x 10-8 *3300m/968
= 0.234 mm
Therefore, the diameter is d= 2r = 2(0.234 mm) = 0.446 mm
(b) To determine the electric field we can use the values for the potential difference across the cable and the length of the cable, where the electric field is inversely proportional to the length of the cable as next
E =V/L
=220/3300m
= 0.066 V/m
ANY DENSE OBJECT
A mathematical equation for density is: Density = mass/volume or D = m/v. If something has a large mass compared to its volume, it has a high density. This is like a set of weights which can be small but heavy. But if an object has a small mass compared to its volume it has a lower density.
Pls give me a brainliest if this helps thx
C, due to stuff and whatnot
Answer:
r= 3.2 cm
Explanation:
Given that
I= 8.7 A
B= 5.4 x 10⁻⁵ T
μo=1.25664 x 10⁻⁶
We know that magnetic filed in wire at a distance r given as
By putting the values
r=0.032 m
r= 3.2 cm
Answer:
Fx = 35.36 N
Fy = 35.36 N
Explanation:
From the question,
The X component of the force is
Fx = Fcos∅.................. Equation 1
Where Fx = X component of the force, F = Force, ∅ = Angle to the horizontal.
Give: F = 50 N, ∅ = 45°
Substitute into equation 1
Fx = 50(cos45°)
Fx = 50(0.7071)
Fx = 35.36 N
Similarly,
For Y component
Fy = Fsin∅
Where F y = Y component
Fy = 50(sin45°)
Fy = 50(0.7071)
Fy = 35.36 N
