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madreJ [45]
3 years ago
11

What is the main function of the mitochondria in a cell?

Physics
2 answers:
Alla [95]3 years ago
7 0

Answer:

Below :)

Explanation:

It is the powerhouse of the cell and creates chemical energy.

Oduvanchick [21]3 years ago
3 0
Answer: Mitochondria are organelles that provide energy to the cell. A mitochondrion is an organelle that contains a nucleus and a small number of mitochondria. Mitochondria are made up of an inner mitochondrial membrane and an outer membrane. A cell contains a number of different types of mitochondria, each with a unique function. Some mitochondria are used to produce

Explanation:
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Early cameras were little more than a box with a pinhole on the side opposite the film. (a) What angular resolution would you ex
ollegr [7]

Answer:

angular resolution = 0.07270° = 1.269 × 10^{-3} rad

greatest distance from the camera = 118.20 m = 0.118 km

Explanation:

given data

diameter = 0.50 mm = 0.5 × 10^{-3} m

distance apart = 15 cm =  15× 10^{-2} m

wavelength λ = 520 nm = 520 × 10^{-9} m

to find out

angular resolution and greatest distance from the camera

solution

first we expression here angular resolution that is

sin θ = \frac{1.22* \lambda }{D}   .......................1

put here value λ is wavelength and d is diameter

we get

sin θ = \frac{1.22*520*10^{-9}}{0.5*10^{-3}}

θ = 0.07270° = 1.269 × 10^{-3} rad

and

distance from camera is calculate here as

θ = \frac{I}{r}    .................2

I = \frac{15*10^{-2}}{1.269*10^{-3}}

I = 118.20 m = 0.118 km

3 0
3 years ago
How long will it take a 400 W engine to lift a 100 N object up to a height of 80 m?
Norma-Jean [14]
<h3>Answer:</h3>

20 seconds

<h3>Explanation:</h3>

<u>We are given</u>;

  • Power of the engine as 400 watts
  • Force as 100 N
  • Distance the object is lifted up as 80 m

We are required to determine the time taken.

  • We need to know that power is the rate of work done

Therefore;

  • Power = Work done ÷ time

But, work done  = Force × distance

  • Therefore; In this case;

Work done = 100 N × 80 m

                  = 8000 Joules

  • Since , Power = Work done ÷ time

Then, time = Work done ÷ Power

Thus;

Time = 8000 J ÷ 400 W

       = 20 s

Therefore, the time taken by the engine to lift the object is 20 seconds

4 0
4 years ago
Read 2 more answers
During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapu
natali 33 [55]

Answer:

110.7 J

Explanation:

Hooke's law is represented by the formula:

F = ke  where F is the force in Newton, K is force constant and e is extension in m

work done = 1/2ke² = 1/2 K ( e² - e₀²) and e₀ is the extension at relaxed length

e₀ =0

work done = 0.5 × 82N/m × (2.70 m)² = 110.7 J

4 0
3 years ago
If manuel weighs 375 n and is holding the box. how much force is the floor exerting on him? (include direction)
denis23 [38]
<span> The box is exerting a force of its weight. So we'll use W=mg (Note: you can use the notation F=ma, doesn't matter since weight IS a force.) We can call g negative since it's directed down towards the earth. 
</span>
<span>W = 5kg(-9.8 m/(s^2)) = -49 N 
</span>
<span>Thus, the box exerts a force of 49N downwards. </span>
4 0
3 years ago
The component of the external magnetic field along the central axis of a 18 18 turn circular coil of radius 27.0 27.0 cm decreas
madreJ [45]

Answer:

Induced current I = 0.44 A

Explanation:

given data

no of turn N = 18

radius  = 27.0 cm

resistance of the coil = 4.50 Ω

time = 2.90 s

magnetic field  = 1.90 T to 0.500 T

solution

we get Induced EMF that is

E = - dφ ÷ dt    ..............1

E = - N × A × (B2 - B1) ÷ Δt

E = N × A × (B1 - B2) ÷ Δt

area of cross section of the coil A = π × r²

area of cross section = 3.14 × (0.27)²

area of cross section = 0.228 m²

Initial magnetic field B1   = 1.9 T

and  

Final magnetic field B2   = 0.5 T

time      Δt = 3.7 s

put here value and we get

E  =  \frac{18 \times 0.228 \times (1.9-0.5)}{2.90}  

E  = 1.98 V

and  

Induced current I is express as

Induced current I = E  ÷ R    ................2

Induced current I = \frac{1.98}{4.5}

Induced current I = 0.44 A

3 0
3 years ago
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