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Oksanka [162]
2 years ago
10

If the atmospheric pressure is 15 lb/in, what is the corresponding downward force on the top of a horizontal square area 5 inche

s on each side ?
Physics
1 answer:
mr Goodwill [35]2 years ago
3 0

The corresponding downward force on the top of a horizontal square is 375 lb.in

<h3>What is pressure?</h3>

The pressure is the amount of force applied per unit area. It is represented as

Pressure p = Force/Area

If the atmospheric pressure is 15 lb/in and  area 5 inches on each side is

A = 5² = 25  in²

The force applied is

15 lb/in = F /25 in²

F = 375 lb.in

Hence, the corresponding downward force on the top of a horizontal square is 375 lb.in

Learn more about pressure.

brainly.com/question/12971272

#SPJ1

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Answer:Eating. Your muscles in your arms and mouth use energy to feed itself. Then your body digest the food which also takes energy.

Sleep. When your tired, you don’t have much energy. It is said that you use more energy while your sleeping. But how do you become energized if you were using even more energy than before?

6 0
2 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
PLEASE ITS AN Emergency IF ITS RIGHT I WILL GIVE BRAINLIEST
n200080 [17]

Answer:

all of those are pisitions

Explanation:

6 0
2 years ago
What are the three ways of answering a scientific question
My name is Ann [436]

Answer:

Let's start by understanding what exactly a scientific question is. A scientific question is a question that may lead to a hypothesis and help us in answering (or figuring out) the reason for some observation. A good scientific question has certain characteristics. It should have some answers (real answers), should be testable.

Here's examples of a few:

Why is that a star?

or

What is that star made of?

Hope this can lead you to the answer you're looking for at least!!

5 0
3 years ago
It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³
Evgesh-ka [11]

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, \rho_o = 800 kg/m³

density of water, \rho_w = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

8 0
3 years ago
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