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PilotLPTM [1.2K]
3 years ago
9

A trait of an organism that helps it survive in its environment is calles

Physics
2 answers:
Oksanka [162]3 years ago
8 0
Hey there!!


Answer: Adaptation


Adaptation is when a living organism evolve to live in a particular habitat.



Hope this helps!!

kompoz [17]3 years ago
7 0
Answer is adaptation. An organism develops a trait over time to help survive in its environment called an adaptation. You could take a giraffe for example. A long time ago giraffes actually had short necks, but now since their food is higher they soon developed a longer neck and this is what we now see in the present. This goes for any artic animal. Polar bears and seals have a white fur adaptation to help them blend in with their environment. A chameleon changes colors in order to hide from predators and sneak up on prey. These are all adaptations
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A form of energy is stored in the bonds between atoms. What is the name for this stored energy?
tester [92]

Answer:

chemical energy

Explanation:

A form of energy is stored in the bonds between atoms is known as chemical energy.

3 0
3 years ago
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You can see stars brighter than magnitude +6 with your naked eye under dark sky conditions, and you can also see the full moon a
Mariana [72]
-6 uh 89 0 -+= 62 uwuwhhiwhiuhwiuhwuihhishishoknsk
6 0
3 years ago
A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming
Monica [59]

Answer:

The deceleration is  a =  - 76.27 m/s^2

Explanation:

From the question we are told that

   The height above  firefighter safety net is H  = 14 \ m

   The length by which the net is stretched is s =  1.8 \ m

   

From the law of energy conservation

    KE_T + PE_T =  KE_B + PE_B

 Where KE_T is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

   and  PE_T is the potential energy of the before jumping  which is mathematically represented at

          PE_T  = mg H

and  KE_B is the kinetic energy of the person just before landing on the safety net  which is mathematically represented at

        KE_B = \frac{1}{2} m v^2

and  PE_B is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

   So the above equation becomes

          mgH =  \frac{1}{2} m v^2

=>           v =  \sqrt{2 gH }

    substituting values

                v =  16.57 m/s

Applying the equation o motion

             v_f =  v  + 2 a s

Now the final velocity is zero because the person comes to rest

      So

         0 = 16.57 + 2 * a * 1.8

            a =  - \frac{16.57^2 }{2 * 1.8}

            a =  - 76.27 m/s^2

         

         

4 0
3 years ago
In a research facility, a person lies on a horizontal platform which floats on a film of air. When the person's heart beats, it
anastassius [24]

Answer: 3.48g

Explanation:

here, we will be using conservation of momentum to solve the problem. i.e the total momentum remains unchanged, unless an external force acts on the system. We'll in thus question, there is no external force acting in the system.

Remember, momentum = mass * velocity, then

mass of blood * velocity of blood = combined mass of subject and pallet * velocity of subject and pallet

Velocity of blood = 56.5cm = 0.565m

mass of blood * 0.565 = 54kg * (0.000063/0.160)

mass of blood * 0.565 = 54 * 0.00039375

mass of blood * 0.565 = 0.001969

mass of blood = 0.00348kg

Thus, the mass of blood that leaves the heart is 3.48g

7 0
3 years ago
Read 2 more answers
an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco
Semenov [28]

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

a_c = \frac{v^2}{r}

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

a_c = \frac{(12\ m/s)^2}{2\ m}

<u>ac = 72 m/s²</u>

<u></u>

Now, the centripetal force applied by the athlete on the hammer will be:

F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)

<u>Fc = 504 N</u>

6 0
2 years ago
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