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VladimirAG [237]
2 years ago
15

PSYCHOLOGY All of our behaviors have been learned. OA True OB False

Physics
1 answer:
Mekhanik [1.2K]2 years ago
6 0

Answer:

A True

the answer is true

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- Virtual

- Upright

- Smaller

Explanation:

I just did it on Edg22 and I got it correct :D

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PLZ HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!<br> What is a front?
Shkiper50 [21]

Answer:

in computer and writing context it is many styles of writing that you can make the text look like. I.E: <em>helllo  </em>hello

Explanation:

A font is the combination of typeface and other qualities, such as size, pitch, and spacing. For example, Times Roman is a typeface that defines the shape of each character. Within Times Roman, however, there are many fonts to choose from -- different sizes, italic, bold, and so on.

Mark me the brainliest plz

4 0
2 years ago
Read 2 more answers
Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest. Which best describes h
meriva

Answer:

8.1km/h Northwest

Explanation:

The 8.1km/h Northwest gives the best description of her distance from start to finish. This distance can be represented in a right angle triangle , this is the hypotenuse which is the longest side of the triangle. If we add 5.7 and 5.8 this gives 11.5km/h compared to 8.1km/h which is a smaller distance and the best.

6 0
2 years ago
Starting from rest, a truck travels in a straight line for 8.0 s with a uniform acceleration of +1.6 m/s2. The driver then appli
lana [24]
Because it should be the total number
4 0
3 years ago
The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh
Aliun [14]

Answer:

The value is      A   = 39315 \  m^2

Explanation:

From the question we are told that

    The velocity which the rover is suppose to land with is  v  =  1 \ m/s

    The  mass of the rover and the parachute is  m  =  2270 \ kg

     The  drag coefficient is  C__{D}}  =  0.5

      The atmospheric density of Earth  is  \rho =  1.2 \  kg/m^3

     The acceleration due to gravity in Mars is  g_m  =  3.689 \  m/s^2

     

Generally the Mars  atmosphere density is mathematically represented as

          \rho_m  =  0.71 *  \rho

=>        \rho_m  =  0.71 *  1.2

=>        \rho_m  = 0.852 \  kg/m^3

Generally the drag force on the rover and the parachute  is mathematically represented as

          F__{D}} =  m  *  g_{m}

=>       F__{D}} =  2270   *  3.689  

=>       F__{D}} =  8374 \ N  

Gnerally this drag force is mathematically represented as

         F__{D}} =   C__{D}} *  A *  \frac{\rho_m * v^2 }{2}

Here A is the frontal area

So  

         A   =  \frac{2 *  F__D }{ C__D}  *  \rho_m  * v^2   }

=>       A   =  \frac{2 * 8374 }{ 0.5 *  0.852    *  1 ^2   }

=>       A   = 39315 \  m^2

8 0
3 years ago
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