The black means that it is a great emitter/absorber of the electromagnetic spectrum. The electromagnetic radiation is reflected of the white and absorbed nurture black meaning that the temperature of the black tarmac increases to that greater the the white
Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be =
..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength =
..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance =
..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
<span>The total
energy stored is the sum of the energy stored in the capacitors. If the
capacitors are series connected
capacitors, then the charging current is the same for both capacitors. This
means that each capacitor stores the same energy and the stored energy is two
times the energy of any of the capacitors.</span>
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
Answer:
20 pig callers
Explanation:
Given that:
A pig caller produced intensity level of a sound = 107 dB
To find how many pig callers required to generate an intensity level of 120 dB;
we have:
120 dB - 107 dB = 13 dB
Taking the logarithm function;

where;
= initial intensity


I = 19.95
I ≅ 20 pig callers