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antoniya [11.8K]

2 years ago

12

Explain how microorganisms are friend and foe.

1 answer:

Helga [31]2 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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Compared to all other forms of energy, wind power is most similar to tidal wave power because _____.

SOVA2 [1]

It is because they are both renewable sources of energy

5 0

3 years ago

Read 2 more answers

The body is subjected to a force of 0,4 N m with a shoulder of 5 cm. What is the magnitude of this force?

igor_vitrenko [27]

Answer:

8N

Explanation:

the body is subjected to pressure equals to force force inverse to the area of that body given that it is in motion

5 0

2 years ago

The following questions consist of a statement and a reason .Answer these questions selecting the appropriate options given belo

Mamont248 [21]

Answer:

<em><u>1)A)</u></em>

<em><u>1)A)2)A)</u></em>

<h3><em><u>Hope it helps you </u></em><em><u>♡</u></em><em><u>♡</u></em></h3>

4 0

2 years ago

what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?

nignag [31]

Using Newton's second law of motion:

F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)

Given: Find: Formula: Solve for m:

F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2

a= 200m/s^2

Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

8 0

3 years ago

A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit

lapo4ka [179]

Answer:

The angle is $\theta = 15.48^o$

Explanation:

From the question we are told that

The distance of the dartboard from the dart is $d = 3.66 \ m$

The time taken is $t = 0.455 \ s$

The horizontal component of the speed of the dart is mathematically represented as

$u_x = ucos \theta$

where u is the the velocity at dart is lunched

so

$distance = velocity \ in \ the\ x-direction * time$

substituting values

$3.66 = ucos \theta * (0.455)$

=> $ucos \theta = 8.04 \ m/s$

From projectile kinematics the time taken by the dart can be mathematically represented as

$t = \frac{2usin \theta }{g}$

=> $usin \theta = \frac{g * t}{2 }$

$usin \theta = \frac{9.8 * 0.455}{2 }$

$usin \theta = 2.23$

=> $tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}$

$\theta = tan^{-1} [0.277]$

$\theta = 15.48^o$

4 0

3 years ago