An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.
The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x
Kg
<h3>
Relationship between Linear and angular speed</h3>
Linear speed is the product of angular speed and the maximum displacement of the particle. That is,
V = Wr
Where
Given that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.
a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed
W = 2
/T
W = (2 x 3.143) / (365.26 x 24)
W = 6.283 / 876624
W = 7.2 x
Rad/hr
The Earth’s average orbital speed V = Wr
V = 7.2 x
x 149.6 x 
V = 107225.5 kilometers per hours.
b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law
M = (4
) / G
M = (4 x 9.8696 x 3.35 x
) / (6.67 x
x 7.68 x
<em>)</em>
<em>M = 1.32 x </em>
/ 51.226
M = 2.58 x
Kg
Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x
Kg
Learn more about Orbital Speed here: brainly.com/question/22247460
#SPJ1
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