To answer this question, we should know the formula for the terminal velocity. The formula is written below:
v = √(2mg/ρAC)
where
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient
Let's determine the mass, m, to be density*volume.
Volume = s³ = (1 cm*1 m/100 cm)³ = 10⁻⁶ m³
m = (1.6×10³ kg/m³)(10⁻⁶ m³) = 1.6×10⁻³ kg
A = (1 cm * 1 m/100 cm)² = 10⁻⁴ m²
v = √(2*1.6×10⁻³ kg*9.81 m/s²/1.6×10³ kg/m³*10⁻⁴ m²*0.8)
<em>v = 0.495 m/s</em>
Answer:
The beat frequency is 30 Hz
Explanation:
Given;
velocity of the two sound waves, v = 343 m/s
wavelength of the first wave, λ₁ = 5.72 m
wavelength of the second wave, λ₂ = 11.44 m
The frequency of the first wave is calculated as follows;
F₁ = v/λ₁
F₁ = 343 / 5.72
F₁ = 59.97 HZ
The frequency of the second wave is calculated as follows;
F₂ = v/λ₂
F₂ = 343 / 11.44
F₂ = 29.98 Hz
The beat frequency is calculated as;
Fb = F₁ - F₂
Fb = 59.97 HZ - 29.98 Hz
Fb = 30 Hz
Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s
The alpha line in the Balmer series is the transition from n=3 to n=2 and with the wavelength of λ=656 nm = 6.56*10^-7 m. To get the frequency we need the formula: v=λ*f where v is the speed of light, λ is the wavelength and f is the frequency, or c=λ*f. c=3*10^8 m/s. To get the frequency: f=c/λ. Now we input the numbers: f=(3*10^8)/(6.56*10^-7)=4.57*10^14 Hz. So the frequency of the light from alpha line is f= 4.57*10^14 Hz.
