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2 years ago

What is the meaning of mito

Physics

2 answers:

marysya [2.9K]2 years ago

7 0

Explanation:

A disorder that occurs when structures that produce energy for a cell malfunction.

A common factor among mitochondrial diseases is that the mitochondria are unable to completely burn food and oxygen to generate energy, which is essential for normal cell function. It's often inherited

Zepler [3.9K]2 years ago

4 0

Answer:

Mito is a legend or untrue story. There are many meanings tbh but that is the one I know. Lol hope this helps

Explanation:

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2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the

frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum = 2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = $\frac{1}{2}$ mv²

v² = 2gh₁

v = √(2gh₁)

from the second equation; s = ut + $\frac{1}{2}$ at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{5}$ mR²) ω²

v = √( $\frac{10}{7}$ gh₁ )

so we substitute √( $\frac{10}{7}$ gh₁ ) for v and t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$ gh₁ ) × √(2h₂/g)

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{3}$ mR²) ω²

v = √( $\frac{6}{5}$ gh₁ )

so we substitute √( $\frac{6}{5}$ gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( $\frac{6}{5}$ gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{5}$ mR²) ω²

v = √( $\frac{10}{7}$ gh₁ )

so we substitute √( $\frac{10}{7}$ gh₁ ) for v and t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$ gh₁ ) × √(2h₂/g)

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0

3 years ago

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air

ValentinkaMS [17]

Answer:

the value of the final pressure is 0.168 atm

Explanation:

Given the data in the question;

Let p₁ be initial pressure, v₁ be initial volume.

After expansion, p₂ is final pressure and v₂ is final volume.

So using the following equations;

p₁v₁ = nRT

p₂v₂ = nRT

hence, p₁v₁ = p₂v₂

we find p₂

p₂ = p₁v₁ / v₂

given that; initial volume v₁ = 0.175 m³, Initial pressure p₁ = 0.350 atm,

final volume v₂ = 0.365 m³

we substitute

p₂ = ( 0.350 atm × 0.175 m³ ) / 0.365 m³

p₂ = 0.06125 atm-m³ / 0.365 m³

p₂ = 0.168 atm

Therefore, the value of the final pressure is 0.168 atm

7 0

2 years ago

Why do electrical devices have resistance

Kamila [148]

As electrons move through the conductor, some collide with atoms, other electrons, or impurities in the metal.

8 0

2 years ago

A proton and an electron are moving in the +x direction in a magnetic field in the +z

Cloud [144]

Answer:

Explanation:

A proton and electron are moving in the positive x direction, this shows that their velocity will be in the positive x direction

V = v•i

Magnetic field Is the positive z direction

B = B•k

A. For proton.

Proton has a positive charge of q

Direction of force on proton

Force is given as

F = q(v×B)

F = q( v•i × B•k)

F = qvB (i×k)

From vectors i×k = -j

F = -qvB •j

Then, for the positive charge, the force will act in the negative direction of the y-axis

B. For electron

Electron has a negative of -q

Direction of force on proton

Force is given as

F = q(v×B)

F = -q( v•i × B•k)

F = -qvB (i×k)

From vectors i×k = -j

F = --qvB •j

F = qvB •j

Then, for the negative charge, the force will act in the positive direction of the y-axis

5 0

3 years ago

To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal

tatiyna

Answer: Add an incline or grade to the road track.

Explanation:
Refer to the figure shown below.

When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.

The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.

At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.

When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).

5 0

2 years ago

What is the meaning of mito​

What is the meaning of mito