Answer:
Explanation:
Assuming no friction between the roller coaster car and the hill, and neglecting air resistance, the kinetic energy the roller coaster car would have at the bottom of the hill would be equal to its gravitational potential energy at the top of the hill, by conservation of energy.
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The question is incomplete. You dis not provide values for A and B. Here is the complete question
Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.
A = 12
B = 18
Answer:
18.5⁰
Explanation:
Angle of incidence i = 12.0 + A
A = 12
= 12.0 + 12
= 14
Refractive index u = 1.10 + B/100
= 1.10 + 18/100
= 1.10 + 0.18
= 1.28
We then find the angle of refraction index u
u = sine i / sin r
u = sine24/sinr
1.28 = sine 24 / sine r
1.28Sine r = sin24
1.28 sine r = 0.4067
Sine r = 0.4067/1.28
r = sine^-1(0.317)
r = 18.481
= 18.5⁰
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
C: Litmus Paper. Red litmus paper turns blue in acids; blue litmus paper turns red in bases
