Answer:
speed and time are Vf = 4.43 m/s and t = 0.45 s
Explanation:
This is a problem of free fall, we have the equations of kinematics
Vf² = Vo² + 2g x
As the object is released the initial velocity is zero, let's look at the final velocity with the equation
Vf = √( 2 g X)
Vf = √(2 9.8 1)
Vf = 4.43 m/s
This is the speed with which it reaches the ground
Having the final speed we can find the time
Vf = Vo + g t
t = Vf / g
t = 4.43 / 9.8
t = 0.45 s
This is the time of fall of the body to touch the ground
Answer:
Negative intrapleural pressure is the correct answer
Explanation:
Intrapleural pressure is more subatmospheric in the uppermost part of the thorax than in the lowermost parts in the standing horse.
Air moves from a region of higher pressure to one of lower pressure. Therefore, for air to be moved into or out of the lungs, a pressure difference between the atmosphere and the alveoli must be established. If there is no pressure difference, no airflow will occur.
Under normal circumstances, inspiration is accomplished by causing alveolar pressure to fall below atmospheric pressure. When the mechanics of breathing are being discussed, atmospheric pressure is conventionally referred to as 0 cm H2O, so lowering alveolar pressure below atmospheric pressure is known as negative-pressure breathing.
As stated in the statement, we will apply energy conservation to solve this problem.
From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as
Where,
m = mass
= initial and final velocity
g = Gravity
h = height
As the mass is tHe same and the final height is zero we have that the expression is now:
Answer:
The periodic table illustrate some of the elements from Hydrogen to Calcium
Answer:
See below explanation
Explanation:
The correspondent chemical reaction for copper carbonate decomposed by heat is:
CuCO₃ (s) → CuO (s) + CO₂ (g)
Considering all molar mass (MM) for each element ( we consider rounded numbers) :
MM CuCO₃ = 123 g/mol
MM CuO = 79 g/mol
MM CO₂ = 44 g/mol
Statement mentions that scientis heated 123.6 g of CuCO₃ (almost a MM), until a black residue is obtained, which weights 79.6 g : this solid residue is formed by CuO, and the remaining mass (approximatelly 44 g) belongs to teh second product, this is, CO₂; as it is a gas compund, it is not certainly included on the solid residue.
So, law of conservation mass is true for this case, since: 123.6 g = 79.6 g + 44 g. As explained, on the solid residue, we don not include the 44 g, which "escaped" from our system, since it is a gas compound (CO₂)
