Answer:
Factors that influence fluid pressure are depth and “ “ of the fluid.
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Answer:
1) m = 0.3137 kg
2a)T_f = -181.7°C
2b) T_f = -1176.97°C
Explanation:
1) We are given;
Length; l = 4.30 cm = 0.043 m
Width; w = 14.0 cm = 0.014 m
height; h = 27.0 cm = 0.027 m
density of gold; ρ = 19.3 × 10³ kg/m³
Formula for the density is known as;
ρ = mass/volume
Thus;
m =ρV
m = 19.3 × 10³ × (lwh)
m = 19.3 × 10³ × (0.043 × 0.014 × 0.027)
m = 0.3137 kg
2a) We are given;
Diameter of brass; L_br = 10 cm
Diameter of aluminum; L_al = 10.01 cm
Now, to some for change in temperature we will use the formula;
L_f = L_i + αL_i(Δt)
Where α is coefficient of expansion.
Now, for the ring to be removed from the rod, the final diameter of the brass has to be same as the aluminium.
Thus;
L_f(brass) = L_f(aluminium)
From table attached, α_brass ≈ 19 × 10^(-6) /°C
Also, α_aluminium ≈ 24 × 10^(-6) /°C
Thus;
L_f(brass) = 10 + (19 × 10^(-6) × 10 × (Δt))
Similarly,
L_f(aluminium) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))
Since L_f(brass) = L_f(aluminium), then;
10 + (19 × 10^(-6) × 10 × (Δt)) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))
Rearranging, we have;
10.01 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))
0.01 = Δt(-50.24 × 10^(-6))
Δt = 0.01/(-50.24 × 10^(-6))
Δt ≈ -199°C
Thus, temperature at which the combination must be cooled to separate the two metals is;
T_f = T_i + Δt
T_f = 17.3 + (-199)
T_f = -181.7°C
2b) Diameter of aluminum is now;
L_al = 10.06 cm
Thus;
10.06 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))
0.06 = Δt(-50.24 × 10^(-6))
Δt = 0.06/(-50.24 × 10^(-6))
Δt = -1194.27°C
T_f = 17.3 + (-1194.27)
T_f = -1176.97°C
