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3 years ago

What cell organelle is necessary for cellular respiration

Physics

2 answers:

podryga [215]3 years ago

4 0

It is the cytoplasm.

jekas [21]3 years ago

3 0

The cell organelle that is necessary for cellular respiration is the mitochondria.

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Which is not a property of a pure substance

Ann [662]

air is a mixture and not a pure substance.

5 0

3 years ago

The energy (such as light and heat) released by the Sun is produced by nuclear reactions in the core of the Sun, which convert h

Yanka [14]

Answer:

<em>the mass of one helium nucleus should be</em> <em>less than the mass of four hydrogen nuclei.</em>

Explanation:

Deep inside the core of the Sun, enough protons can collide into each other with enough speed that they stick together to form a helium nucleus and generate a tremendous amount of energy at the same time. This process is called nuclear fusion.

The mass-to-energy conversion is described by Einstein's famous equation:

E = mc2, or, in words, energy equals mass times the square of the velocity of light. Because the velocity of light is a very large number, this equation says that lots of energy can be gained from using up a modest amount of mass.

Photons In the proton-proton chain reaction, hydrogen nuclei are converted to helium nuclei through a number of intermediates. The reactions produce high-energy photons (gamma rays) that move through the "radiative layer" surrounding the core. This layer takes up 60 percent of the radius of the Sun. It takes a million years for energy to get through this layer into the "convective layer", because the photons are constantly intercepted, absorbed and re-emitted. In the core, the helium nuclei make up 62% of the mass (the rest is still hydrogen). The radiative and convective layers have about 72% hydrogen, 26% helium, and 2% heavier elements (by mass). The energy produced by fusion is then transported to the solar surface and emitted as light or ejected as high-energy particles.

3 0

3 years ago

A ray of light travels from a glass-to-liquid interface at an angle of 35.0º. Indices of refraction for the glass and liquid are

UkoKoshka [18]

Answer:

The angle of refraction for the ray moving through the liquid is = 32.3°

Explanation:

Refractive index of liquid (n₁/n₂) = sini/sinr

∴ n₁/n₂ = sini/sinr ................ equation 1

n₁ = index of refraction for glass, n₂ = index of refraction for liquid

Where i = incident angle of the first medium, r = angle of refraction or angle in the second medium.

Since the light ray is traveling from glass - to - liquid, the first medium is glass and the second medium is liquid. and the refractive index will be that liquid with respect to glass.

using the equation,

n₁/n₂ = sini/sinr

i = 35° , n₁ = 1.52, n₂= 1.63

Making sinr the subject of the equation above,

sinr = sini/(n₁/n₂)

sinr = sin35(1.52)/1.63

sinr =0.574(1.52)/1.63

sinr = 0.535

Taking the sin inverse of both side of the equation

sin⁻¹(sinr) = sin⁻¹(0.535)

∴ r = 32.3°

The angle of refraction for the ray moving through the liquid is = 32.3°

The right option is (b). 32.3°

5 0

4 years ago

Chapter 05, Problem 15 Multiple-Concept Example 7 and Concept Simulation 5.2 review the concepts that play a role in this proble

Llana [10]

Answer:

Explanation:

The question relates to motion on a circular path .

Let the radius of the circular path be R .

The centripetal force for circular motion is provided by frictional force

frictional force is equal to μmg , where μ is coefficient of friction and mg is weight

Equating cenrtipetal force and frictionl force in the case of car A

mv² / R = μmg

R = v² /μg

= 26.8 x 26.8 / .335 x 9.8

= 218.77 m

In case of moton of car B

mv² / R = μmg

v² = μRg

= .683 x 218.77x 9.8

= 1464.35

v = 38.26 m /s .

3 0

3 years ago

The "lead" in pencils is a graphite composition with a Young’s modulus of about 1×1010N/m21×1010⁢N/m2. Calculate the change in l

Tanya [424]

Answer:

b) 0.1 mm

Explanation:

Given that

E= 1 x 10¹⁰ N/m²

F= 4 N

d= 0.5 mm

L = 60 mm

We know that elongation due to force F given as

$\Delta L=\dfrac{FL}{AE}$

$\Delta L=\dfrac{FL}{\dfrac{\pi d^2}{4}\times E}$

$\Delta L=\dfrac{4\times 60}{\dfrac{\pi \times 0.5^2}{4}\times 10^4}$

ΔL = 0.12 mm

Therefore the answer is -

b) 0.1 mm

6 0

3 years ago