Which claim is true of all matter?

A technician is checking refrigerant system pressures. Both high- and low-side service ports are located on the A/C compressor.

konstantin123 [22]

Answer:

Two major causes are outline bellow

1. The presence of air in the system

2. Clogged condenser

Explanation:

1. The presence of air in the system

One of the causes that have been established in relation to high compressor discharge pressure is the presence of air in the system. When this takes place, your best solution is to recharge the system.

2. Clogged condenser

Another is a clogged condenser in which case you will need to clean the condenser so that it will function properly. When you happen to spot that the discharge valve is closed and it is causing high discharge pressure on the compressor, you can solve that easily by opening the valve

5 0

4 years ago

An adult dog can sit, rollover, and bark when needs to potty outside. these are a type of what behavior.

ss7ja [257]

Training behaviors..........................................

5 0

4 years ago

element X has two isotopes: X-27 and x-29. x-27 has an atomic mass of 26.975 and a relative abundance of 82.33%, and X-29 has an

GalinKa [24]

Answer:

27.34 (no unit)

Explanation:

26.975*82.33%+29.018*17.67%

=27.34

7 0

3 years ago

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

You often hear people, particularly TV announcers, talk about a "high rate of speed." What do you think they mean? According to

NikAS [45]

TV announcer intend to mean "chnage of speed or velocity" by "High rate of speed" and in Physics domain it would mean "acceleration"

Explanation:

It is common to observe TV announcer saying certain events were occurring “with a high rate of speed”. By saying this they intend to mean that the event was rapid in its occurrence. It can also mean that the change in speed of the happening was very rapid/fast.

However, the same terms connote altogether a different expression in Physics domains. Speed is a scalar quantity with no direction. Hence most of the times speed mean velocity when the direction is also provided. “high rate of speed” would mean a change of velocity per unit time which is acceleration. Hence in Physics domain, the term would stand for acceleration.

3 0

3 years ago