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Helen [10]
3 years ago
12

What are two roles of bacteria in a nitrogen cycle

Physics
2 answers:
evablogger [386]3 years ago
7 0

Answer:Nitrogen-fixing bacteria convert free nitrogen gas into nitrogen compounds. ... Bacterias that are decomposers recycle nitrogen compounds in the soil by breaking down animal wastes and dead plants and animals. 3. Other bacteria break down nitrogen compounds and release free nitrogen back into the air.

Explanation:

sweet-ann [11.9K]3 years ago
3 0

Answer:

Nitrogen-fixing bacteria in the soil and within the root nodules of some plants convert nitrogen gas in the atmosphere to ammonia. Nitrifying bacteria convert ammonia to nitrites or nitrates.

Explanation:

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An observer is standing next to the tracks, watching a train approach. The train travels at 30 m/s and blows its whistle at 8,00
SSSSS [86.1K]

7351.35Hz

f0= v-Vo/v-Vs × FSA

= 340-0 /340+30 ×8000

= 340/370× 8000

= 7351.35hz

7 0
3 years ago
When a voltage difference is applied to a piece of metal wire, a current flows through it. if this metal wire is now replaced wi
Nikitich [7]

The resistance of the cylindrical wire is R=\frac{\rho l}{A}.

Here R is the resistance, l is the length of the wire and A is the area of cross section. Since the wire is cylindrical A=\frac{\pi d^2}{4} .

Comparing two wires,

R_1=\frac{\rho_1 l}{A_1} \\ R_2=\frac{\rho_2 l}{A_2}

Dividing the above 2 equations,

\frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{A_2 }{A_1}  \\ \frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{d_2^2 }{d_1^2}  \\

Since d_2=2d_1

The above ratio is

\frac{R_1}{R_2}=\frac{1.68(10^{-8})  }{1.59(10^{-8}) } (4)\\ \frac{R_1}{R_2}=4.2264

We also have,

\frac{E/R_1}{E/R_2} =\frac{I_1}{I_2} \\ I_2=\frac{R_1}{R_2}I_1 \\ I_2=4.23I_1

The current through the Silver wire will be 4.23 times the current through the original wire.

8 0
3 years ago
You hold your physics textbook in your hand. (Assume that no other objects are in contact with the book.)
Jet001 [13]

Answer:

1) the correct answer is b and d

2) For force b its direction is vertical up

   for the force d its direction is vertical down

3) the correct answers are: a, c and d

4) Force a is vertical down , force c is vertical up and force d is vertical down

5) the correct answer which is b

Explanation:

In this exercise it is asked to identify the forces, fundamentally on the free there are the forces of gravity and the support force of the hand, with these facts we answer the questions

1) the correct answer is b and d

the hand acts on the book with a contact force and the Earth acts on the book with the force of gravity.

2) For force b its direction is vertical up

   for the force d its direction is vertical down

3) The forces on the hand are the weight of the book. The force of gravity due to the mass of the hand. As the hand is in balance, there must be a force applied by the arm to keep the hand in position; assuming the hand is in the air, if the hand is resting on the floor the force of the floor on the hand can perform this function

therefore the correct answers are: a, c and d

4) Force a is vertical down

    force c is vertical up

   force d is vertical down

5) The action and reaction forces are forces of equal magnitude, each applied to one of the bodies, we have

* the force of the hand on the free and its reaction the force of the book on the hand

* The force of the Earth on the book and the hand, giving the weight of each one and the relationship is the force of the book and the hand on the Earth

the correct answer which is b

8 0
3 years ago
The
guapka [62]
The NUCLEUS is the center of the atom. it contains protons and NEUTRONS

protons have a POSITIVE charge

neutrons have a NEUTRAL charge

electrons have a NEGATIVE charge
6 0
3 years ago
How far will it go, given that the coefficient of kinetic friction is 0.10 and the push imparts an initial speed of 3.9 m/s ?
Akimi4 [234]

Answer:19.5 m

Explanation:

Given

coefficient of kinetic Friction \mu _k=0.10

Initial speed u=3.9\ m/s

Friction is present so it tries to stop to the object and stops it completely after moving certain distance let say s

maximum deceleration provided by friction is

a_{max}=\mu _k\cdot g

a_{max}=0.1\times 3.9=0.39\ m/s^2

using equation of motion

v^2-u^2=2 as

where v=final\ velocity

u=initial velocity

a=acceleration

s=displacement

(0)-(3.9)^2=2\times (-0.39)\times s

s=19.5\ m

6 0
3 years ago
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