If we assume also that the temperature of the air does not change, we can use Boyle's Law:
p₁V₁ = p₂V₂
Now, we know:
p₁ = 100kPa
V₂ = 100cm³ (the volume of the tyre)
V₁ = 120cm³ (becuse the air is contained inside the tyre AND the pump)
We can solve for p₂:
p₂ = (p₁V₁)/V₂
= (100×120)/100
= 120kPa
Therefore your answer is: 120kPa
Answer:
P₂ = 392720.38 Pa = 392.72 kPa
Explanation:
Given
D₁ = 5 cm = 0.05 m
D₂ = 10 cm = 0.10 m
v₁ = 8 m/s
P₁ = 380 kPa = 380000 Pa
α = 1.06
ρ = 1000 kg/m³
g = 9.8 m/s²
We can use the following formula
(P₁ / (ρg)) + α*(V₁² / (2g)) + z₁ = (P₂ / (ρg)) + α*(V₂² / (2g)) + z₂ + +hL
knowing that z₁ = z₂ we have
(P₁ / (ρg)) + α*(V₁² / (2g)) = (P₂ / (ρg)) + α*(V₂² / (2g)) + +hL <em> (I)
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Where
V₂ can be obtained as follows
V₁*A₁ = V₂*A₂ ⇒ V₁*( π* D₁² / 4) = V₂*( π* D₂² / 4)
⇒ V₂ = V₁*(D₁² / D₂²) = (8 m/s)* ((0.05 m)² / (0.10 m)²)
⇒ V₂ = 2 m/s
and
hL is a head loss factor: hL = α*(1 - (D₁² / D₂²))²*v₁² / (2*g)
⇒ hL = (1.06)*(1 – ((0.05 m) ² / (0.10 m)²))²*(8 m/s)² / (2*9.8 m/s²)
⇒ hL = 1.9469 m
Finally we get P₂ using the equation <em>(I)
</em>
⇒ P₂ = P₁ - ((V₂² - V₁²)* α*ρ / 2) – (ρ*g* hL)
⇒ P₂ = 380000 Pa - (((2 m/s)² - (8 m/s)²)*(1.06)*(1000 kg/m³) / 2) – (1000 kg/m³*9.8 m/s²*1.9469 m)
⇒ P₂ = 392720.38 Pa = 392.72 kPa
Answer:
I) Heat capacity of ice (Sensible, ice), II) Enthalpy of fusion of water (Latent, fusion) and III) Heat capacity of water (Sensible, water).
Explanation:
At standard pressure, water has a fusion point and boiling point of 273.15 K and 373.15 K, respectively. Then, the total change in enthalpy is the sum of two sensible indicators and one latent indicator. That is:
The needed physical constants are: Heat capacity of ice (Sensible, ice), Enthalpy of fusion of water (Latent, fusion) and Heat capacity of water (Sensible, water).
Answer:
true
Explanation:
for example assume you are setting in a moving bus and when someone see you from the ground you are in motion but for some who is with you in the bus you are not in motion.