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antoniya [11.8K]
3 years ago
12

Explain how microorganisms are friend and foe.​

Physics
1 answer:
Helga [31]3 years ago
8 0

Answer:

Because they are beneficial

Explanation:

There are microorganisms in our large intestine that synthesis vitamins and allow them to be absorbed into the bloodstream. However a tiny minority are pathogens (disease-causing agents). These pathogens often called germs or bugs, are a threat to all life forms.

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Why is it that the weight of an object weighing 1N air, weighs more when immersed in water ?
Anni [7]
There is no "why", because that's not what happens.  The truth is
exactly the opposite. 

Whatever the weight of a solid object is in air, that weight will appear
to be LESS when the object is immersed in water.

The object is lifted by a force equal to the weight of the fluid it displaces.
It displaces the same amount of air or water, and any amount of water
weighs more than the same amount of air.  So the force that lifts the
object in water is greater than the force that lifts it in air, and the object
appears to weigh less in the water.
4 0
3 years ago
Writing a 4 to 5 sentence paragraph, identify YOUR stance on whether more Nuclear Reactors should be built here in Michigan. (1s
prisoha [69]
Thank You for giving me the points ^_^
4 0
3 years ago
A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

3 0
2 years ago
Linear expansivity of steel
Anuta_ua [19.1K]

<em>Steel: 11.0 – 12.5</em>

<em>T̶e̶t̶s̶u̶t̶e̶t̶s̶u̶ ̶T̶e̶t̶s̶u̶t̶e̶t̶s̶u̶</em>

Thanks,

<em>Deku ❤</em>

5 0
3 years ago
A ball is launched from ground level at 20 m/s at an angle of 40° above the
DedPeter [7]

(a) The ball's height <em>y</em> at time <em>t</em> is given by

<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :

0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )

<em>t</em> = 0   or   (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 <em>g t</em>

<em>t</em> = (40 m/s) sin(40º) / <em>g</em>

<em>t</em> ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So

0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>

where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :

<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)

<em>y</em> ≈ 8.4 m

8 0
3 years ago
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