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joja [24]
2 years ago
12

You look in the sky and see two jetliners that you know are of equal size, yet one appears to be much larger. Because of your kn

owledge of __________, you will assume that the smaller jetliner is farther away
Physics
1 answer:
Ksju [112]2 years ago
8 0

Answer:

Because of the knowledge of <u>relative size</u>, it will be assumed that the smaller jetliner is farther away.

Explanation:

According to the theory of relative size, the distance that an object has to the viewing individual affects the perception of the individual regarding the size of the object.

As stated in this case, one of the jetliners is farther away from the other. Therefore, even if the jets are of equal size, the one that is at a greater distance is perceived to be smaller as it is at a greater viewing range. The one that is nearer to the individual seems bigger in comparison to the one farther away due to a closer viewing range.

Therefore, the jet that is nearer appears larger.

To know more about relative size, refer to:

brainly.com/question/19998265

#SPJ4

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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
A 23.0 kg iron weightlifting plate has a volume of 2920 cm3 . what is the density of the iron plate in g/cm3?
yanalaym [24]
The first thing you should know for this case is that density is defined as the quotient between mass and volume:
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 In addition, you should keep in mind the following conversion:
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 Substituting the values we have:
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 answer
 the density of the iron plate is 7.88 g / cm ^ 3
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A 2.8-kg physics cart is moving forward with a speed of 45 cm/s. A 1.9-kg brick is dropped from rest and lands on the cart. The
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