El unico que se es el del oxígeno: o2
Answer:
his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.
determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.
a)0.7956kg/s
b)5.437 × 10⁻³m²
Explanation:
The concepts related to the change of mass flow for both entry and exit is applied
The general formula is defined by
Where,
values are divided by inlet(1) and outlet(2) by
PART A) Applying the flow equation
PART B) For the exit area we need to arrange the equation in function of Area, that is
We have that the the scenarios the flux is behavior is as follows:
- Here The -ve charge placed outside the surface will not affect the total flux charge.
- We have that the flux remains constant.
- Here with the <em>negative </em>charge outside, the flux will decrease.
- This will not affect net flux change, whereas if it where inside.It will increase flux.
Generally for each change
A positive charge is placed just inside the surface:
Here The -ve Charge placed outside the surface will not affect the total flux charge.
The charge is moved of-center by an amount equal to half the radius of the spherical surface.
We have that the flux remains constant.
A negative charge is placed just outside the surface.
Here with the <em>negative </em>charge outside, the flux will decrease.
A <u>positive </u><em>charge </em>is placed just outside the surface.
This will not affect net flux change, whereas if it where inside.It will increase flux.
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Complete Question
A positive point charge is placed at the center of an imaginary spherical surface. As a result, there is some total electric flux outward through the surface. For each change listed, how would the total outward flux though the surface change?.
- A positive charge is placed just inside the surface .
- The charge is moved of-center by an amount equal to half the radius of the spherical surface .
- A negative charge is placed just outside the surface.
- A positive charge is placed just outside the surface.
Answer:
I_total = L² (m + M / 3)
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
It is appreciated that it is a scalar quantity, for which it is additive, in this case the system is formed by two bodies and the moment of inertia must be the sum of each moment of inertia with respect to the same axis of rotation.
The moment of inertia of a bar with respect to an axis that passes through ends is
I_bar = 1/3 M L²
The moment of inertia of a particle is
I_part = m x²
We have to assume the point where the particle sticks to the bar, suppose it sticks to the end
x = L
Total moment of inertia is the sum of these two moments of inertia
I_total = I_bar + I_particule
I_total = 1/3 M L² + m L²
I_total = L² (m + M / 3)
Answer:
A. 1.6 × 105 joules
Explanation:
As per the question, the data given in the question is as follows
Number of masses = 2
Each weightage =
Speed = 12.5 meters/ second
Based on the above information, the approximate kinetic energy after the collision is
A perfectly elastic collision is described as one where the collision does not cause any loss of kinetic energy.
So we sum the kinetic energy of each kind of system which is given below:
Kinetic energy is
= 156250 J