Answer:
P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,
Explanation:
This problem of fluid mechanics let's start with the continuity equation to find the speed of water output
Q = A v
v = Q / A
The area of a circle is
A = π r² = π d² / 4
Let's look at the speeds at each point
v₁ = Q / A₁ = Q 4 /π d₁²
v₁ = 10 4 /π 0.5²
v₁ = 50.93 m / s
v₂ = Q / A₂
v₂ = 10 4 /π 0.25²
v₂ = 203.72 m / s
Now we can use Bernoulli's equation in the colon
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear
P₁ = P2 + ½ rho (v₂² - v₁²)
P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)
P₁ = 1.013 10⁵ + 2.205 10⁷
P₁ = 2.215 10⁷ Pa
la definicion de presion es
P₁ = F₁/A₁
F₁ = P₁ A₁
F₁ = 2.215 10⁷ pi d₁²/4
F₁ = 2.215 10⁷ pi 0.5²/4
F₁ = 4.3 106 N
When two different air masses meet, a boundary is formed. the boundary between two air masses is called a front. weather at a front is usually cloudy and stormy. there at four different fronts: cold, warm, stationary, and occluded
Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
Answer:
Energy expenditure in K cals/min = 10 K cals /min (approximately)
Explanation:
As we know
Energy expenditure in Kcal/min= METs x 3.5 x Body weight (kg) / 200
Given is METs=7.6
Weight of Jazz= 172lb=78.02kg
putting the values in formula,
Energy expenditure in K cals/min= 7.6 x 3.5 x 78.02 / 200
=10.38 K cals /min
=10 K cals /min (approximately)
Therefore, Energy expenditure in K cals/min by Jazz will be approximately 10 K cals /min
Dropping a bouncy ball and stretching a rubber ban.
