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3 years ago

PLEASE HELP! URGENT Explain how the forces need to change so an aeroplane can land

Physics

1 answer:

dangina [55]3 years ago

6 0

Answer:

it changes by taking the air from below the plane and curving it to the top causing draw wich slows it down then the weight that pulls it down to land.

please put me brainliest

Explanation:

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What is the lupac name H3C – CHT the H₂ - CH - CH2 CH₂ CH₂ CH3

dimulka [17.4K]

Answer:??

Explanation:

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2 years ago

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

Answer:

Thunderbird is 995.157 meters behind the Mercedes

Explanation:

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$

initial velocity $u= 71 m/s$ distance

Distance covered in the slowing down phase = $250 m$

$v^2-u^2=2as$

$a= \frac {(v^2-u^2)}{2s}$

$a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2$

$v=u+at$

$t= \frac {(v-u)}{a}$

$= \frac {(0-71)}{(-10.082)}=7.042 s$

$t_1=7.042 s$

The car is in the pit stop for 5s $t_2=5 s$

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

$a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2$

$v=u+at$

$t= \frac{(v-u)}{a}$

$t= \frac{(71-0)}{6.81}= 10.425 s$

$t_3=10.425 s$

total time= $t_1 +t_2+t_3=7.042+5+10.425=22.467 s$

Distance covered by the Mercedes Benz during this time is given by $s=vt=71 \times 22.467= 1595.157 m$

Distance covered by the Thunderbird during this time= $250+350=600 m$

Difference between distance covered by the Mercedes and Thunderbird

= $1595.157-600=995.157 m$

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0

3 years ago

A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in

lilavasa [31]

Answer:

Explanation:

The change is as follows

P₁ V₁ to 3P₁, V₁ ( constt volume ) --- first process

3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process

In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁

P₁V₁ = n R T₁ , n is no of moles of gas enclosed.

nRT₁ = P₁V₁

Heat added at constant volume = n Cv ( 3T₁ - T₁)

= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)

= 10/3 x nRT₁

= 10/3x P₁V₁

In the second process, Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁

Heat added at constant pressure in second case

= n Cp ( 15T₁ - 3T₁)

= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)

= 28 x nRT₁

= 28 P₁V₁

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3 years ago

What is the best description of the end result of chemical bonding for most

forsale [732]

A. Getting a full set of valence electrons

Explanation:

The best description of the end result of chemical bonding for most atoms is the getting of a full set of valence electrons.

Atoms reacts with one another in order to complete valence electronic shell.

The valence electron shell is the outermost energy level of an atom.
It is from this energy level that electrons are lost or gained to form bonds.
All atoms wants to be like the noble gases whose valence electronic shell is completely filled up
This is the crux of chemical bonding
The attraction that is produced from the interaction leads to bond formation

learn more:

Chemical bond brainly.com/question/10903097

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3 years ago

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Vedmedyk [2.9K]

B density will increase

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3 years ago

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