What is the magnification formula?

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

In order for a ball to move upward, the initial velocity of the ball must be greater than _____.

ch4aika [34]

Answer:

The answer is zero please Give me Brainly

Explanation:

5 0

2 years ago

Match the type of heat transfer with its description

VikaD [51]

Answer:

1. Convection

2. Radiation

3. Conduction

Hope this helps!

Explanation:

5 0

3 years ago

Between which two points of a concave mirror should an object be placed to obtain a magnificati of -3

kobusy [5.1K]

Answer:

When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.

When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.

Explanation:

8 0

3 years ago

suppose that you had an ohmmeter and a mystery component that could be either an inductor or a capacitor. How could you use the

sukhopar [10]

Answer:

To determine the mystery component we will connect the mystery component to a DC voltage source, then I will measure the resistance of the component with the use of Ohmmeter, the value of the resistance of the mystery component will determine what the mystery component is

if the resistance > 1( very high ) then component is a capacitor

if the resistance = 0 then component is an inductor

Explanation:

To determine the mystery component we will connect the mystery component to a DC voltage source, then I will measure the resistance of the component with the use of Ohmmeter, the value of the resistance of the mystery component will determine what the mystery component is

if the resistance > 1( very high ) then component is a capacitor

if the resistance = 0 then component is an inductor

8 0

2 years ago