Answer:
The range of powers is 
Explanation:
From the question we are told that
The far point of the left eye is 
The near point of the left eye is 
The near point with the glasses on is 
From these parameter we can see that with the glass on that for near point the
Object distance would be 
Image distance would be 
To obtain the focal length we would apply the lens formula which is mathematically represented as

substituting values


converting to meters


Generally the power of the lens is mathematically represented as

Substituting values


From these parameter we can see that with the glass on that for far point the
Object distance would be 
Image distance would be 
To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

substituting values


converting to meters

Generally the power of the lens is mathematically represented as

Substituting values


This implies that the range of powers of the lens in his glass is

The answer is : C ) air,water,and the tank glass.
-Hope this helps.
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