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77julia77 [94]
3 years ago
12

Identify the different parts of the face ​

Physics
1 answer:
Vaselesa [24]3 years ago
7 0

Answer:

head

four head

hair

eyes

ears

nose

lips

chick

chin

eyebrow

eyelid

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What changes dead plants and animals into ammonia compounds?
lapo4ka [179]

Nitrogen fixing bacteria changes dead plants and animals into ammonia compounds.

<h2>What is nitrogen fixation?</h2>

Atmospheric nitrogen is converted into nitrogen oxides by the action of lightning, which helps its incorporation into the soil.

<h3>Characteristics of Nitrogen fixing bacteria</h3>

  • Nitrogen is fixed by these bacteria and other prokaryotes through various metabolic processes, which convert it into different usable compounds, such as ammonia (NH3) and ammonium ion (NH4+).

  • These microorganisms can be found in soil and water, or as plant symbionts.

Therefore, we can conclude that nitrogen fixing bacteria fix nitrogen from the air, that is, they originate soluble compounds by plants, such as ammonia.

Learn more about nitrogen fixation here: brainly.com/question/14726009

7 0
2 years ago
For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo
Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}

so that its period is T=3.644\,\frac{\mathrm s}{\mathrm{rev}} (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration \Delta t. Denote by \theta the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ

\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ

\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ

We can then compute the magnitude of the velocity vector differences \Delta\vec v for each time interval by using the law of cosines:

|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta

\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}

and in turn we find the magnitude of the average acceleration vectors to be

\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0
3 years ago
The magnetic force of a magnet is stronger at its poles than in the middle.<br> true or false
earnstyle [38]

The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole when compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center.

8 0
3 years ago
Read 2 more answers
10 POINTS + BRAINLIEST TO FIRST CORRECT ANSWER! :)
enyata [817]

Answer:

The 2nd option

Explanation:

It slows down the transfer of thermal energy from inside to outside the coat.

3 0
3 years ago
Which one is it<br> Plz help me :)
Gennadij [26K]

Answer:

5 I think will be none of the above and 6 could be all of the above

8 0
3 years ago
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