All categories

3 years ago

The production of carbon dioxide might be an indicator of what process

Physics

2 answers:

disa [49]3 years ago

6 0

Its a proces of combustion "burning"humans produce carbon dioxide,rocks<plants whatever is burning energy

aalyn [17]3 years ago

3 0

The carbon dioxide is produced by two methods naturally, which is either photosynthesis or by decomposition. The indicator thus for the carbon dioxide production be the decomposition most likely.

<u>Explanation:</u>

The carbon dioxide is produced by the process of decomposition and photosynthesis both. But in the process of photosynthesis it is released as a by product which then consumed in the process of respiration. Therefore if there is ample production, it indicated of decomposition. The decomposition is the decaying process of organic matter.

You might be interested in

The position vector of a particle of mass 1.65 kg as a function of time is given by = (6.00 î + 4.15 t ĵ), where is in meters an

SashulF [63]

Answer:

L = 41.09 Kg m2 / s The angular momentum does not depend on the time

Explanation:

The definition of angular momentum is

L = r x p

Where blacks indicate vectors

Let's apply this definition our case. Linear momentum

p = m v

Let's replace

L = m r x v

The given function is

x = 6.00 i ^ + 4.15 t j ^

We look for speed

v = dx / dt

v = 0 + 4.15 j ^

To evaluate the angular momentum one of the best ways is to use determinants

$L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]$

L = m 6 4.15 k ^

The other products give zero

Let's calculate

L = 1.65 6 4.15 k ^

L = 41.09 Kg m2 / s

The angular momentum does not depend on the time

7 0

3 years ago

A wind turbine works by slowing the air that passes its blades and converting much of the extracted kinetic energy to electric e

ddd [48]

Answer:

2649600 Joules

Explanation:

Efficiency = 40%

m = Mass of air = 92000 kg

v = Velocity of wind = 12 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 92000\times 12^2\\\Rightarrow K=6624000\ J$

The kinetic energy of the wind is 6624000 Joules

The wind turbine extracts 40% of the kinetic energy of the wind

$E=0.4\times K\\\Rightarrow E=0.4\times 6624000\\\Rightarrow E=2649600\ J$

The energy extracted by the turbine every second is 2649600 Joules

8 0

3 years ago

At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th

Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

3 years ago

transmission electron microscopes that use high-energy electrons accelerated over a range from 40.0 to 100 kv are employed in ma

Gekata [30.6K]

The spatial limitations in Picometer for the given range of electrons would be around 50 picometers.

What is a transmission electron microscope?

A transmission electron microscope (TEM) is a type of microscope that uses a beam of high-energy electrons to produce detailed images of the structure of materials at the atomic or molecular scale. TEMs work by passing a focused beam of electrons through a thin sample and collecting the transmitted electrons on a fluorescent screen or an electronic detector. The interaction of the sample with the electrons results in the formation of an image that can be magnified and displayed on a computer monitor. TEMs are widely used in the fields of materials science, biology, and nanotechnology and can provide information about the structure, composition, and properties of materials with a high level and resolution.

According to the problem:

The spatial resolution of a transmission electron microscope (TEM) is determined by the size of the electron probe, which is directly related to the energy of the electrons. The higher the energy of the electrons is, the smaller the size of the probe is and the higher the spatial resolution.

At the lower end of the energy range of 40.0 kV, the spatial resolution of the TEM would be on the order of hundreds of nanometers. At the higher end of the range (100 kV), the spatial resolution would be on the order of tens of nanometers.

In general, TEMs with electron energy in the range of 40-100 kV are capable of resolving details down to around 50 picometers (pm). However, the actual spatial resolution will depend on various factors, such as the quality of the electron optics, the stability of the electron beam, and the sample preparation.

It's worth noting that TEMs with even higher electron energies (up to several hundred kV) are available, which can achieve spatial resolutions down to the sub-angstrom level (less than 0.1 pm). However, these instruments are much more expensive and complex to operate than TEMs with lower electron energies.

To know more about de broglie wavelength, visit:

brainly.com/question/17295250

#SPJ4

7 0

1 year ago