At a high school science fair, Connor won first place for his replica of the Golden Gate Bridge. Connor liked the project so muc
1 answer:
You might be interested in
Answer:
Explanation:
Given data:
flow rate = 10 gallon per minute = 0.0223 ft^3/sec
diameter = 0.75 inch
we know discharge is given as
Q = VA
solve for velocity V = \frac{Q}{A}[/tex]
V = 7.27 ft/sec
we know that Reynold number
calculate the ratio to determine the fanning friction f
from moody diagram f value corresonding to Re and is 0.037
for horizontal pipe
where 1.94 slug/ft^3is density of water
Answer:
T_{f} = 90.07998 ° C
Explanation:
This is a calorimetry process where the heat given by the Te is absorbed by the air at room temperature (T₀ = 25ºC) with a specific heat of 1,009 J / kg ºC, we assume that the amount of Tea in the cup is V₀ = 100 ml. The bottle being thermally insulated does not intervene in the process
Qc = -Qb
M (T₁ -
) = m
(T_{f}-T₀)
Where M is the mass of Tea that remains after taking out the cup, the density of Te is the density of water plus the solids dissolved in them, the approximate values are from 1020 to 1200 kg / m³, for this calculation we use 1100 kg / m³
ρ = m / V
V = 1000 -100 = 900 ml
V = 0.900 l (1 m3 / 1000 l) = 0.900 10⁻³ m³
V_air = 0.100 l = 0.1 10⁻³ m³
Tea Mass
M = ρ V_te
M = 1100 0.9 10⁻³
M = 0.990 kg
Air mass
m = ρ _air V_air
m = 1.225 0.1 10⁻³
m = 0.1225 10⁻³ kg
(m c_{e_air} + M c_{e_Te}) T_{f}. = M c_{e_Te} T1 - m c_{e_air} T₀
T_{f} = (M c_{e_Te} T₁ - m c_{e_air} T₀) / (m c_{e_air} + M c_{e_Te})
Let's calculate
T_{f} = (0.990 1100 90.08– 0.1225 10⁻³ 1.225 25) / (0.1225 10⁻³ 1.225 + 0.990 1100)
T_{f} = (98097.12 -3.75 10⁻³) / (0.15 10⁻³ +1089)
T_{f} = 98097.11 / 1089.0002
T_{f} = 90.07998 ° C
This temperature decrease is very small and cannot be measured
Answer:
The density of the unknown liquid is 1.025 kg/m³ (considering the density of the water as 1.000 kg/m³)
Explanation:
The hydrometer works by the Archimedes principle. The cylinder floats in the liquid because the hydrostatic thrust is equal to the weight force. This means:
If we measure 2 fluids, the weight of the hydrometer is the same, so:
If the original watermark height is 12.3cm (H₁) and the mark for water has risen 0.3 cm above the unknown liquid–air interface, the height of the unknown liquid mark is 12cm (H₂). Therefore: