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2 answers:
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Answer:
Part 1
Part 2
Explanation:
Given that
Diameter,d=1 μm
Length ,l=2 μm
As we know that volume of cylinder given as
Surface area,A
A=π d l
Part 1
Part 2
The frequency of middle C on a string is
f = 261.6 Hz.
The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m
The length of the string is L = 1 m.
Let T = the tension in the string (N).
The velocity of the standing wave is
In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N
Answer: 136.9 N (nearest tenth)
f = 261.6 Hz.
The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m
The length of the string is L = 1 m.
Let T = the tension in the string (N).
The velocity of the standing wave is
In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N
Answer: 136.9 N (nearest tenth)
In the above problem, we need to find mass of the second child, so that the Center of Mass remains at the origin( pivot).
CM= m1r1+m2r2/m1+m2
0= 20*-2+16*r2/20+16
r2= 40/16
r2= +2.5 m
CM= m1r1+m2r2/m1+m2
0= 20*-2+16*r2/20+16
r2= 40/16
r2= +2.5 m