Explanation:
It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.
The charge per unit length of the wire is 
and the net charge per unit length is 
.
We know that there exist zero electric field inside the metal cylinder.
(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let 
are the charge per unit length on the inner and outer surfaces of the cylinder.
For inner surface,
For outer surface,
(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :
Hence, this is the required solution.
(a) 
The frequency of a wave is given by:
where
v is the wave's speed
is the wavelength
For the red laser light in this problem, we have
(speed of light)
Substituting,
(b) 427.6 nm
The wavelength of the wave in the glass is given by
where
is the original wavelength of the wave in air
n = 1.48 is the refractive index of glass
Substituting into the formula,
(c) 
The speed of the wave in the glass is given by
where
is the original speed of the wave in air
n = 1.48 is the refractive index of glass
Substituting into the formula,
Answer:
Explanation: Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outermost (valence) electrons, atoms can fill up their outer electron shell and gain stability.
7.5 x 10⁻¹¹m. An electromagnetic wave of frecuency 4.0 x 10¹⁸Hz has a wavelength of 7.5 x 10⁻¹¹m.
Wavelength is the distance traveled by a periodic disturbance that propagates through a medium in a certain time interval. The wavelength, also known as the space period, is the inverse of the frequency. The wavelength is usually represented by the Greek letter λ.
λ = v/f. Where v is the speed of propagation of the wave, and "f" is the frequency.
An electromagnetic wave has a frecuency of 4.0 x 10 ¹⁸Hz and the speed of light is 3.0 x 10⁸ m/s. So:
λ = (3.0 x 10⁸ m/s)/(4.0 x 10¹⁸ Hz)
λ = 7.5 x 10⁻¹¹m
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v = 
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ = 
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
