Answer:
Glucose and Oxygen
Explanation:
Cellular respiration is the process whereby cells derives energy by the use of glucose and oxygen.
Organisms that use cellular respiration to produce their energy are known as heterotophs. They derive the glucose from food materials obtained from plant sources. They use the oxygen from the environment to liberate energy from the glucose obtained from feeding on plant materials.
Cellular respiration can be simply expressed as shown below:
GLUCOSE + OXYGEN → CO₂ + H₂O + ATP
The reactants are glucose and oxygen.
The products are CO₂, water and ATP
The acceleration of the runner in the given time is 2.06m/s².
Given the data in the question;
Since the runner begins from rest,
- Initial velocity;
- Final velocity;
- Time elapsed;
Acceleration of the runner; 
<h3>Velocity and Acceleration</h3>
Velocity is the speed at which an object moves in a particular direction.
Acceleration is simply the rate of change of the velocity of a particle or object with respect to time. Now, we can see the relationship from the First Equation of Motion
Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.
To determine the acceleration of the runner, we substitute our given values into the equation above.
Therefore, the acceleration of the runner in the given time is 2.06m/s².
Learn more about Equations of Motion: brainly.com/question/18486505
F=ma
F= 4x1.2
F= 4.8 N
F= 4gsin30 - Friction
Friction= 19.6 - 4.8 N
Friction= 14.8 N
Friction= u x 4gcos30
14.8 / 4gcos30 = u
u= 0.43596...
u= 0.44
coefficient is 0.44
Answer:
P₂ = 138.88 10³ Pa
Explanation:
This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation
Let's start by looking for the top speed
Q = A₁ v₁ = A₂ v₂
We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)
v₂ = A₁ / A₂ v₁
They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)
v₂ = 2 1.20
v₂ = 2.40 m / s
Now let's write the Bernoulli equation
P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂
Let's clear the pressure at point 2
P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)
we put our reference system at the lowest point
y₁ - y₂ = -20 cm
Let's calculate
P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)
P₂ = 143 103 - 2,160 103 - 1,960 103
P₂ = 138.88 10³ Pa
Answer:
the rock hit the board you go flying because you board stops but you have energy still going so there for you go flying
Explanation:
