Explanation:
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The short answer (and the long one for that matter) is physical properties of chemicals. If you are being marked by a machine, likely the answer is going to be physical properties.
Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
<u>FOR BULB 1:</u>
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
<u>FOR BULB 2:</u>
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
<u>A₁/A₂ = 0.44</u>
Answer:
11.07Hz
Explanation:
Check the attachment for diagram of the standing wave in question.
Formula for calculating the fundamental frequency Fo in strings is V/2L where;
V is the velocity of the wave in string
L is the length of the string which is expressed as a function of its wavelength.
The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)
Therefore L = 1.5λ
If L = 3.0m
1.5λ = 3.0m
λ = 3/1.5
λ = 2m
Also;
V = √T/m where;
T is the tension = 0.98N
m is the mass per unit length = 2.0g = 0.002kg
V = √0.98/0.002
V = √490
V = 22.14m/s
Fo = V/2L (for string)
Fo = 22.14/2(3)
Fo = 22.14/6
Fo = 3.69Hz
Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo
Frequency of the wave = 3×3.69
Frequency of the wave = 11.07Hz