Answer:

Explanation:
From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.
Hence, the material dispersion is 
Now, using the pulse spread formula:


Thus, the pulse spreading as a result of material dispersion is:
Answer:
1.5 m/s
Explanation:
Momentum is conserved and conservation of momentum is
p₁ + p₂ = p'₁ + p'₂
or
m₁v₁ + m₂v₂ = m₁v'₁ + m₂v'₂
In our problem, after collision v'₁ will be equal to v'₂.
Since objects are identical m₁ = m₂
m(v₁+ v₂) = 2m x v'₁
(2m/s + 1m/s) = 2v'₁
v'₁ = v'₂ = 1.5 m/s
Answer:
The frequency would double.
Explanation:
Given:
Speed of wave (v) = constant.
Frequency of wave initially (f₁) = 2 Hz
Initial wavelength of the wave (λ₁) = 1 m
Final wavelength of the wave (λ₂) = 0.5 m
Final frequency of the wave (f₂) = ?
We know that the product of wavelength and frequency of the wave is equal to the speed of the wave.
Therefore, framing in equation form, we have:
Wavelength × Frequency = Speed

It is given that speed of the wave remains the same. So, the product must always be a constant.
Therefore,

Now, plug in the given values and solve for 'f₂'. This gives,

Therefore, the final frequency is 4 Hz which is double of the initial frequency.
f₂ = 2f₁ = 2 × 2 = 4 Hz
So, the second option is correct.
Before the impact, let the velocity of the baseball was v m/s.
After being hit by the bat its velocity is -2v
So, change in velocity, Deltav=v-(-2v)=3v
Acceleration is defined as the rate of change in velocity, i.e. actual change in velocity divided by the time taken to change it. Time taken to change velocity is the time of actual contact of the bat and ball, i.e. 0.31 s.
a=(Deltav)/(Deltat)
=(3v)/0.37
Therefore, a/v=3/0.31=9.7 s^-1
So, the ratio of acceleration of the baseball to its original velocity is 9.7.
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV =
mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV =
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
K_p = ½ m v_e²
K_p =
9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² =
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = 
v/c= 2.33 10⁻⁴