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vagabundo [1.1K]
3 years ago
5

What changes in airplane longitudinal control must be made to maintain altitude while the airspeed is being decreased?

Physics
1 answer:
garik1379 [7]3 years ago
8 0

Explanation:

The  changes can be made in airplane longitudinal control to maintain altitude while the airspeed is being decreased is

We can increase the angle of attack this would compensate for the decreasing lift. As the angle of attack directly controls the distribution of pressure on the wings. Moreover, increase in angle of attack will also cause the drag to increase.

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devlian [24]

Answer:

https://youtu.be/ymHHdoCGJOU

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3 years ago
A chemical change can be represented by ____________ changing into _____________ in a ________________.
Harrizon [31]

Answer:

I don't know but I explain some of it as best as I can

Explanation:

Substances that start a chemical reaction are called reactants. Substances that are produced in the reaction are called products. Reactants and products can be elements or compounds. Bonds break in the reactants and new bonds form in the products.

3 0
3 years ago
The weight of a body floating in a liquid is​
djverab [1.8K]

Answer:

The apparent weight is the weight of the body minus the weight of the liquid displaced. The body will float only when both the weights are same. In this case, the given body of weight W is floating and hence the apparent weight is zero.

4 0
3 years ago
Read 2 more answers
A petrol engine that transforms 1000J of chemical potential energy into 300J of kinetic energy, and 700J into wasted heat and so
sergey [27]

Answer:

Efficiency = 30% = 0.3

Explanation:

The general formula for efficiency of a device is given as:

Efficiency = (Desired Output/ Input) * 100%

Here, in our case, we have a petrol engine as a device. So, we analyze it for the efficiency calculations. Here, we have:

Chemical Potential Energy = 1000 J

Kinetic Energy = 300 J

Heat and Sound Energy = 700 J

Now, we know that the desired output of a car or the purpose of a car is to provide Kinetic energy, while all other forms of energy such as heat and sound energies are produced as waste. And the chemical energy is provided to car as input, in form of fuel. Therefore,

Input = Chemical Potential Energy = 1000 J

Desired Output = Kinetic Energy = 300 J

Therefore,

Efficiency = (300 J/1000 J) * 100%

<u>Efficiency = 30% = 0.3</u>

4 0
3 years ago
Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85% of the of the entering water is e
Natalka [10]

Answer:

a)

i) v'=\frac{17}{20}                                   ii) \frac{m_v}{m_f-m_v} =\frac{17}{83}

b) m_r=752963.55\ kg

Explanation:

Given:

fraction of water in wet sugar of m kg by mass, m'_w=\frac{1}{5} \times m

% of water evapourated from the total water after passing through the evapourator, m'_v=85\%

a)

amount of wet sugar fed to the evapourator, m_f=100\ kg

Now the mass of water present in the fed amount of sugar:

m_w=\frac{1}{5} \times m_f

m_w=\frac{100}{5}

m_w=20\ kg

Now the amount of water leaving from this total amount of water after passing through the evapourator:

m_v=m_v' \times m_w

m_v=\frac{85}{100} \times 20

m_v=17\ kg

i)

So, the fraction of of water leaving the evapourator:

v'=\frac{m_v}{m_w}

v'=\frac{17}{20}

ii)

Now the ratio of kg water vaporized/kg wet sugar leaving the evaporator.:

\frac{m_v}{m_f-m_v} =\frac{17}{100-17}

\frac{m_v}{m_f-m_v} =\frac{17}{83}

b)

amount of sugar fed per day, m_f=907185\ kg

<u>Now the mass of water in the given amount of sugar per day:</u>

m_w=\frac{m_f}{5}

m_w=\frac{907185}{5}

m_w=181437\ kg

Mass of water vapourized after passing through the evaporator:

m_v=\frac{85}{100}\times m_w

m_v=\frac{85}{100}\times 181437

m_v=154221.45\ kg

Now the mass of water still remaining in the sugar:

m_r=m_w-m_v

m_r=907185-154221.45

m_r=752963.55\ kg

is the mass of extra water to be evapourated.

8 0
3 years ago
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