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Drupady [299]
3 years ago
15

Which part of a cell releases energy that the cell uses? Cell wall Mitochondria Nucleus Vacuole

Physics
2 answers:
MrMuchimi3 years ago
7 0
Answer is...

Mitochondria...
It is the power house of cell...



HOPE IT HELPS YOU '_'
andrew-mc [135]3 years ago
6 0

Answer:

Mitochondria

Explanation:

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The sum of the kinetic and potential energies of a system of objects is conserved: Group of answer choices only when no external
uysha [10]

The sum of the kinetic and potential energies of a system of objects is conserved only when no external force acts on the objects.

<h3>Conservation of mechanical energy</h3>

The principle of conservation of mechanical energy states that the total mechanical energy of an isolated system (absence of external force) is always constant.

M.A = P.E + K.E

where;

P.E is potential energy

K.E is kinetic energy

Thus, the sum of the kinetic and potential energies of a system of objects is conserved only when no external force acts on the objects.

Learn more about conservation of mechanical energy here: brainly.com/question/24443465

7 0
2 years ago
Some children are underfed and often have no place to sleep. According to
jeka94

Answer:

Physiological needs

Explanation:

3 0
3 years ago
What is the median of 5 12 7 15 3 6 1
xxMikexx [17]

Answer:

6

Explanation:

The median is found by listing the numbers in order numerically and finding the one that is right in the middle. In this case you have 1, 3, 5, 6, 7, 12, 15 where the middle term is 6.

7 0
3 years ago
A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el
GuDViN [60]

Answer:

7.2\cdot 10^{-19} J

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

\Delta U = q \Delta V

where

q is the magnitude of the charge of the particle

\Delta V is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J

- the potential difference is

\Delta V=4.50 V

So, the change in electrical potential energy is

\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J

7 0
3 years ago
2. Two toy cars are involved in a race. Car A has mass m while car B has mass 2m. a. The two cars have the same force applied to
ArbitrLikvidat [17]

Answer:

a) The kinetic energy of the two cars is the same

the moment of car 2 is greater than the moment of car 1

b)  the kinetic energy of car 1 is greater than that of car 2

the moment of the two cars is the same

Explanation:

a) to know the kinetic energy of each car, we must find the speed, use Newton's second law to find the acceleration

Car 1

     F = m a

    a = F / m

Let's use kinematics to find the velocity after x = 1 m

       v² = v₀² + 2 a x

The initial speed is zero

       v = √ (2 F/m  x)

For the distance of x = 1 m

        v₁ = √ (2 F / m)

Car 2

      F = 2m a

      a = F / 2m

      v² = 2 a x

      v = √ (F/m  x)

 For x = 1 m

       v₂ = √(F / m)

Let's calculate the kinetic energy of each car

Car 1

      K₁ = ½ m v₁²

      K₁ = ½ m 2F / m

      K₁ = F

Car 2

      K₂ = ½ 2m v₂²

      K₂ = ½ 2m F / m

      K₂ = F

The kinetic energy of the two cars is the same

Let's calculate the moment

Car 1

   P₁ = m v₁

   P₁ = m √ (2F / m)

Car 2

    P₂ = 2m v²

    P₂ = 2m √(F / m)

We see that the moment of car 2 is greater than the moment of car 1

b) in this part the force is applied by t = 10 s

Acceleration is the same, let's find the speed

Car1

          v = v₀ + a t

          v = F / m t

          v₁ = F / m 10

Car 2

           v₂ = F / 2m 10

           v₂ = F / m 5

Let's calculate the kinetic energy of each car

Car 1

           K₁ = ½ m v₁²

           K₁ = ½ m (F / m 10)²

           K₁ = 50 F² / m

Car2

         K₂ = ½ 2m v₂²

         K₂ = m (F / m 5)²

         K₂ = 25 F² / m

In this case we see that the kinetic energy of car 1 is greater than that of car 2

Let's calculate the moment

Car 1

         P₁ = m v₁

         P₁ = m F / m 10

         P₁ = 10 F

 

Car 2

        P₂ = 2m v₂

        P₂ = 2m F / m 5

        P₂ = 10 F

In this case the moment of the two cars is the same

7 0
3 years ago
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