Answer:
P₁ = 2.3506 10⁵ Pa
Explanation:
For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
A₁ v₁ = A₂ v₂
Let's look for the areas
r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm
r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm
A₁ = π r₁²
A₁ = π 1.125²
A₁ = 3,976 cm²
A₂ = π r₂²
A₂ = π 0.1²
A₂ = 0.0452 cm²
Now with the continuity equation we can look for the speed of water inside the hose
v₁ = v₂ A₂ / A₁
v₁ = 11.2 0.0452 / 3.976
v₁ = 0.1273 m / s
Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)
P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂
P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂
Let's calculate
P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25
P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴
P₁ = 2.3506 10⁵ Pa
The μs between the clock and floor is 650(M*g) and the μk between the clock and the floor is 560(M*g)
Answer:
He calculated his average speed.
Explanation:
Waldo calculated an average of his speed in 2 hours rather than his speed whilst driving which means instead of doing 50/1 hour = 50 mph, he did 50/2hours which is how he got 25 mph. Hope this helps!
Answer:
Explanation:
Given
mass of ice 
Final temperature of liquid 
Specific heat of water 
Latent heat of fusion 
Latent heat of vaporization 
Suppose M is the mass of steam at 
Heat required to melt ice and convert it to water at 
Heat released by steam
and 
must be equal as the heat gained by ice is equal to Heat released by steam
![\Rightarrow M=\dfrac{m[L+c\times T_f]}{L_v+c(100-T_f)}](https://tex.z-dn.net/?f=%5CRightarrow%20M%3D%5Cdfrac%7Bm%5BL%2Bc%5Ctimes%20T_f%5D%7D%7BL_v%2Bc%28100-T_f%29%7D)
![\Rightarrow M=\dfrac{119[333\times 10^3+4186\times 57]}{2256\times 10^3+4186\times (100-57)}](https://tex.z-dn.net/?f=%5CRightarrow%20M%3D%5Cdfrac%7B119%5B333%5Ctimes%2010%5E3%2B4186%5Ctimes%2057%5D%7D%7B2256%5Ctimes%2010%5E3%2B4186%5Ctimes%20%28100-57%29%7D)
