Answer:
346.66 Hz
Explanation:
= Length of string which is unfingered = l
= Length of string which is vibrate when fingered = 
= Unfingered frequency = 260 Hz
= Fingered frequency
Frequency is inversely proportional to length

So,

The frequency of the fingered string is 346.66 Hz
Answer:
182 to 3 s.f
Explanation:
Workdone for an adiabatic process is given as
W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)
where γ = ratio of specific heats. For carbon dioxide, γ = 1.28
For an adiabatic process
P₁V₁ʸ = P₂V₂ʸ = K
K = P₁V₁ʸ
We need to calculate the P₁ using ideal gas equation
P₁V₁ = mRT₁
P₁ = (mRT₁/V₁)
m = 2.80 g = 0.0028 kg
R = 188.92 J/kg.K
T₁ = 27°C = 300 K
V₁ = 500 cm³ = 0.0005 m³
P₁ = (0.0028)(188.92)(300)/0.0005
P₁ = 317385.6 Pa
K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89
W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)
V₁ = 0.0005 m³
V₂ = 2.10 dm³ = 0.002 m³
1 - γ = 1 - 1.28 = - 0.28
W =
18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)
W = -67.47 (5.698 - 8.4)
W = 182.3 = 182 to 3 s.f
Answer:
The volume of the submerged part of her body is 
Explanation:
Let's define the buoyant force acting on a submerged object.
In a submerged object acts a buoyant force which can be calculated as :
ρ.V.g
Where ''B'' is the buoyant force
Where ''ρ'' is the density of the fluid
Where ''V'' is the submerged volume of the object
Where ''g'' is the acceleration due to gravity
Because the girl is floating we can state that the weight of the girl is equal to the buoyant force.
We can write :
(I)
Where ''W'' is weight
⇒ If we consider ρ =
(water density) and
and replacing this values in the equation (I) ⇒


ρ.V.g = 610N
(II)
The force unit ''N'' (Newton) is defined as

Using this in the equation (II) :



We find that the volume of the submerged part of her body is 