<h3><u>Answer;</u></h3>
Electron cloud
<h3><u>Explanation;</u></h3>
- <em><u>An atom is the smallest particle of an element that can take part in a chemical reaction. Atom is made up of two parts ; that is the nucleus and the electron cloud. The nucleus contain subatomic particles; protons and neutrons, while the electron cloud contains the electrons.</u></em>
- <em><u>The electron cloud is the largest part of the atom and is mostly an empty space. Most of an atom is a cloud of electrons surrounding a space called the nucleus with tiny protons and neutrons.</u></em>
Answer:
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Answer:
R₁ = 50.77 Ω
Explanation:
Since, we know that:
Electric Power = P = VI
but from Ohm's Law:
V = IR
(or) I = V/R
Therefore,
P = V²/R
(OR) R = V²/P
where,
V = Battery Voltage
R = Resistance of combination
FOR SERIES COMBINATION:
R = Rs = (57 V)²/48 W
Rs = 67.69 Ω
but, we know that:
Rs = R₁ + R₂
R₁ + R₂ = 67.69 Ω
R₁ = 67.69 Ω - R₂ __________ eqn (1)
FOR PARALLEL COMBINATION:
R = Rp = (57 V)²/256 W
Rp = 12.69 Ω
but, we know that:
Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω
using eqn (1) and value of R₁ + R₂, we get
Rp = 12.69 = R₂(67.69 - R₂)/67.69
859.08 = 67.69 R₂ - R₂²
R₂² - 67.69 R₂ + 859.08 = 0
Solving this quadratic equation we get the answers:
Either, R₂ = 50.76 Ω
Either, R₂ = 16.92 Ω
Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,
<u>R₂ = 16.92 Ω</u>
using this value in eqn (1), we get:
R₁ = 67.69 Ω - 16.92 Ω
<u>R₁ = 50.77 Ω</u>
Answer:
835.29 Hz
Explanation:
When moving towards the source of sound, frequency will be given by
f*=f(vd+v)/v
Where f is the freqiency of the source, vd is the driving speed, v is the speed of sound in air, f* is the inkown frequency when moving forward.
Substituting 800 Hz for f, 340 m/s for v and 15 m/s for vd then
f*=800(15+340)/340=835.29411764704 Hz
Rounded off, the frequency is approximately 835.29 Hz
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2
Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2
Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2
Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2
Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]
Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm
Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm