9 tenths plus two fifths

The temperature of air changes from 0 to 18°C while its velocity changes from zero to a final velocity, and its elevation change

irina [24]

Answer:

For the air:

Final Velocity 160.77m/s

Final Elevation 1,317.43m

the Internal, Kinetic, and Potential Energy changes will be equal.

Explanation:

In principle we know the following:

Internal Energy: is defined as the energy contained within a system (in terms of thermodynamics). It only accounts for any energy changes due to the internal system (thus any outside forces/changes are not accounted for). In S.I. is defined as $U=mC_{V}\Delta T$ where $m$ is the mass (kg), $C_{V}$ is a specific constant-volume (kJ/kg°C) and $\Delta T$ is the Temperature change in °C.

Kinetic Energy: denotes the work done on an object (of given mass $m$ ) so that the object at rest, can accelerate to reach a final velocity. In S.I. is defined as $K=\frac{1}{2}mv^2$ where $v$ is the velocity of the object in (m/s).
Potential Energy: denotes the energy occupied by an object (of given mass $m$ ) due to its position with respect to another object. In S.I. is defined as $P=mgh$ , where $g$ is the gravity constant equal to $9,81m/s^2$ and $h$ is the elevation (meters).

Note: The Internal energy is unaffected by the Kinetic and Potential Energies.

Given Information:

Temperature Change 0°C → 18°C ( thus $\Delta T=18$ °C )
Object velocity we shall call it $v_{o}$ and $v_{f}$ , for initial and final, respectively. Here we also know that $v_{o}=0m/s^2$
Object elevation we shall call it $h_{o}$ and $h_{f}$ , for initial and final, respectively. Here we also know that $h_{o}= 0m$

∴ We are trying to find $v_{f}$ and $h_{f}$ of the air where $U$ , $K$ and $P$ are equal.

Lets look at the change in Energy for each.

Step 1: Change in Kinetic Energy=Change in Internal Energy

$\Delta E_{K}=\Delta U\\\frac{1}{2}m{v_{f}}^2- \frac{1}{2}m{v_{o}}^2=mC_{V}\Delta T$

Here we recall that $v_{o}=0m/s^2$ and mass $m$ is the same everywhere. Thus we have:

$\frac{1}{2}m{v_{f}}^2=mC_{V}\Delta T$

$\frac{1}{2} {v_{f}}^2=C_{V}\Delta T\\ {v_{f}}^2=2C_{V}\Delta T\\ v_{f}=\sqrt{2C_{V}\Delta T}$ Eqn(1)

Step 2: Change in Potential Energy=Change in Internal Energy

$\Delta E_{P}=\Delta U\\mgh_{f}-mgh_{o}=mC_{V}\Delta T$

Here we recall that $h_{o}=0m/s^2$ and mass $m$ is the same everywhere. Thus we have:

$mg(h_{f}-h_{o})=mC_{V}\Delta T\\gh_{f}=C_{V}\Delta T\\$

$h_{f}=\frac{C_{V}\Delta T}{g}$ Eqn(2).

Finally by plugging the known values in Eqns (1) and (2) we obtain:

$v_{f}=\sqrt{2*718*18}=160.77m/s$

$h_{f}=\frac{718*18}{9.81}=1,317.43m$

Thus we can conclude that for the air final velocity $v_{f}=160.77m/s$ and final elevation $h_{f}=1,317.43m$ the internal, kinetic, and potential energy changes will be equal.

3 0

3 years ago

Cara menghitung daya dorong

Arlecino [84]

Thrust adalah kekuatan atau dorongan. Ketika sistem mendorong atau mempercepat massa dalam satu arah, ada dorong (force) hanya besar dalam arah yang berlawanan. Dalam matematika dan fisika, ini dijelaskan oleh hukum kedua dan ketiga Isaac Newton. Dorong digunakan untuk menggambarkan seberapa baik jenis mesin berjalan. Hal ini dapat digunakan untuk berbagai jenis mesin seperti mesin roket, pesawat (baling-baling) mesin, dan mesin jet.
Thrust diukur dalam "pon dorong" di AS dan di Newton dalam sistem metrik. 4.45 Newton dorong sama dengan 1 pon dorong. Satu pon dorong adalah berapa banyak dorong yang dibutuhkan untuk menjaga satu pon objek tak bergerak melawan gaya gravitasi di bumi.

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4 years ago

When light is incident normally on the interface between two transparent optical media, the intensity of the reflected light is

Alex777 [14]

i need to ask sum so oycixgixigdtdidit

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3 years ago

Newton's 3rd Law is based on what type of forces?

NikAS [45]

Answer: His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.

Explanation:

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3 years ago

Read 2 more answers

Initially, the translational rms speed of a molecule of an ideal gas is 349 m/s. The pressure and volume of this gas are kept co

Vedmedyk [2.9K]

Answer:

The final translational rms speed is 247.51 m/s

Explanation:

The translational rms speed, v= 349 m/s.

Translational rms speed of a molecule of an ideal gas is: