All categories

3 years ago

Does air pressure increase or decrease with an increase in altitude is the rate of change constant

1 answer:

7 0

The rate of pressure decreases with an increase in altitude is not constant. Rather, pressure decreases rapidly near Earth's surface and more gradually at greater heights.

You might be interested in

What element from the periodic table rhymes with extreme

jonny [76]

Halite or sulfur or gold or silver

8 0

3 years ago

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

Which of the following statements is TRUE about updating the exposure control plan?

iren2701 [21]

Statements that are true as regards exposure control plan and its updating are;

Updates must have the reflection of changes in tasks as well in procedures.

Updates must reflect changes in positions that affect occupational exposure.

Updates must have the cost of PPE that is needed and necessary to reduce exposure

An exposure control plan can be regarded as the framework for compliance between the employer and the workers.

This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.

This plan gives hope to workers in term of protection when working with their Employer.

There are some elements that is associated with Exposure Control Plan, and theses are;

Health hazards as well as risk that is attributed to each product in the worksite.
Statement of purpose.
procedures and practices in a written form
Responsibilities from the Manager, CEO, designated resources and employer.

Therefore, exposure control plan is avenue to protect workers from bloodborne pathogens.

brainly.com/question/1203927?referrer=searchResults

3 0

2 years ago

A plane is going at a speed of 300 km/h at 63 W of N. The wind hits the plane at a direction of 65 km/h at 52 S of E. What is th

katovenus [111]

It’s around the g force so it’s gonna be around 54 km/h

3 0

1 year ago

A volleyball player bumps a ball across a net with the velocity and angle shown below. What is the maximum height of the ball?

Marrrta [24]

Answer:

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

h = u² sin²θ/2g m

Where,

g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

h = 18² · sin²60 / 2 x 9.8

= 18² x 0.75 / 2 x 9.8

= 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

6 0

3 years ago