Answer:
(a) 0.71 mm
(b) 0.158 cubic cm
Explanation:
The width of one wire is the diameter of the wire.
(a) Let the diameter of each wire is d.
So, 10 d = 14.2 mm
d = 1.42 mm
radius of each wire, r = d/2 = 1.42/2 = 0.71 mm
(b) Length, L = 10 cm
The volume of the single wire is given by

Answer:

Explanation:
From the question we are told that
Mass 
Height 
Generally the equation for velocity before impact is is is mathematically given by



Generally the equation for Kinetic Energy is is mathematically given by




Explanation:
its hard to explain its very complex but its so they can function properly
Answer:
(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz
(b) 250 m/s
(c) 1250 N
(d) Positive x-direction
(e) 6.00 m/s
(f) 0.0365 m
Explanation:
(a) The standard form of the wave is:
y = A cos ((2πf) t ± (2π/λ) x)
where A is the amplitude, f is the frequency, and λ is the wavelength.
If the x term has a positive coefficient, the wave moves to the left.
If the x term has a negative coefficient, the wave moves to the right.
Therefore:
A = 0.0800 m
2π/λ = 0.300 m⁻¹
λ = 20.9 m
2πf = 75.0 rad/s
f = 11.9 Hz
(b) Velocity is wavelength times frequency.
v = λf
v = (20.9 m) (11.9 Hz)
v = 250 m/s
(c) The tension is:
T = v²ρ
where ρ is the mass per unit length.
T = (250 m/s)² (0.0200 kg/m)
T = 1250 N
(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).
(e) The maximum transverse speed is Aω.
(0.0800 m) (75.0 rad/s)
6.00 m/s
(f) Plug in the values and find y.
y = (0.0800 m) cos((75.0 rad/s) (2.00 s) − (0.300 m⁻¹) (1.00 m))
y = 0.0365 m
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
