Answer:
r = 41.1 10⁹ m
Explanation:
For this exercise we use the equilibrium condition, that is, we look for the point where the forces are equal
∑ F = 0
F (Earth- probe) - F (Mars- probe) = 0
F (Earth- probe) = F (Mars- probe)
Let's use the equation of universal grace, let's measure the distance from the earth, to have a reference system
the distance from Earth to the probe is R (Earth-probe) = r
the distance from Mars to the probe is R (Mars -probe) = D - r
where D is the distance between Earth and Mars
M_earth (D-r)² = M_Mars r²
(D-r) = 
r
r ( 
) = D
r = 
We look for the values in tables
D = 54.6 10⁹ m (minimum)
M_earth = 5.98 10²⁴ kg
M_Marte = 6.42 10²³ kg = 0.642 10²⁴ kg
let's calculate
r = 54.6 10⁹ / (1 + √(0.642/5.98) )
r = 41.1 10⁹ m
Answer:
Explanation:
We shall apply Doppler's effect of sound .
speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at 6 m/s .
apparent frequency = 
V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .
Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f
apparent frequency = 
= 
So m = 346 , n = 330 .
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
