All categories

3 years ago

What creates the changes in step 1,2,3 of the nitrogen cycle

1 answer:

6 0

1-fixation ( Bacteria Converts nitrogen to ammonium so plants can use it )
2-nitrification ( bacteria changes ammonium to nitrates and plants )
3 - Assimilation (plants absorb nitrates it is then used for Chlorophyll..)

You might be interested in

a football field is 150 meters long. a football player take 30 seconds to run the entire length of the field. what was his speed

andreyandreev [35.5K]

his speed is : FAST AS ALL GET OUT

8 0

3 years ago

Read 2 more answers

You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand at 30 m/s. What is the kinetic energy (KE) of the volley

FinnZ [79.3K]

awnser :

Explanation:

8 0

2 years ago

Read 2 more answers

Describe a situation where you can be traveling at a low speed but have an extremely high velocity

Reil [10]

Answer:

Cruising at 35,000 feet in an airliner, straight toward the east,

at 500 miles per hour

Explanation:

3 0

2 years ago

Read 2 more answers

A. praised

krok68 [10]

Answer:

14. scolded

15. supporter

16. i am not sure but I think it is removes

17. comfortable

18. outdated

19. drink

20. it can be fantastic or wealthy

8 0

2 years ago

Read 2 more answers

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

3 years ago