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3 years ago

What are the two ost important processes in the oxygen in and out of the atmospher

Physics

2 answers:

torisob [31]3 years ago

5 0

Photosynthesis and respiration

BaLLatris [955]3 years ago

3 0

Decomposing - When plants and animals die, they decompose. This process uses up oxygen and releases carbon dioxide. Rusting - This is also called oxidation. When things rust they use up oxygen.
$hope \: i \: helped \\ \\ \\ \\ \\ \\ darklynx$

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To enhance the effective surface, and hence the chemical reaction rate, catalytic surfaces often take the form of porous solids.

marysya [2.9K]

Answer:

See solution with all the conditions considered. A gaseous mixture of A and B for which species A is chemically consumed at the catalytic surface.

The total pore reaction rate is stated below and it can be inferred by applying the bellow analogy.

4 0

3 years ago

Find the charge on the capacitor in an LRC-series circuit when L = 1 2 h, R = 10 Ω, C = 0.01 f, E(t) = 50 V, q(0) = 1 C, and i(0

alex41 [277]

Answer:

Explanation:

Given that,

In an LRC circuit

L = 1/2h

R = 10 Ω,

C = 0.01 f

E(t) = 50 V,

q(0) = 1 C, and

i(0) = 0 A.

q(t) = C

We can to fine the charge after a long time, let say t→∞

The Kirchoff second law for the system is

L•dq²/dt + R•dq/dt + q/C = E(t)

Divide through by L

dq²/dt + R/L •dq/dt + q/LC = E(t)/L

Now inserting the values of R, L, C and E

dq²/dt+10/½ •dq/dt +q/½×0.01=50/½

dq²/dt + 20•dq/dt + 200q = 100

Let solve the differential equation

First : homogenous solution

Using D operator

D² + 20D +200 = 0

Solving the quadratic equation using formula method

D = (-b±√b²-4ac)/2a

D = (-20±√20²-4×1×200) /2

D = (-20±√400-800)/2

D = (-20±20•i)/2

D = -10±10•i

So we have a complex solution

Then, the complementary solution is

q(t) = e(-10t)[ Acos10t + BSin10t]

A and B are constant

Let find the particular solution using the method of undetermine coefficient

Let assume particular solution of

q(t) = C, I.e q(t) Is a constant

So, inserting this into the equation below

dq²/dt + 20•dq/dt + 200q = 100

200q = 100

q = 100/200

q = ½

Then, the particular solution is ½

So, the total solution is the sum of particular solution and complementary solution

q = e(-10t)[ Acos10t + BSin10t] + ½

Using the initial conditions

q(0) = 1

1 = e(0) [ACos0 + BSin0] +½

1 = A+½

A = ½

Also i(0) = 0

I(t) = q'(t)

Then,

q'(t) = -10•e(-10t)[ Acos10t + BSin10t] + e(-10t)[ -10Asin10t + 10BCos10t]

0 = -10e(0) [ ACos0 + BSin0] + e(0)[-10ASin0 +10BCos0]

0 = -10(A) + 10B

A=B=½

So the general equation becomes

q(t) = e(-10t)[ ½cos10t + ½Sin10t] + ½

So, as t→∞, the aspect of e(-10t) become zero

So the charge stabilizes at q = ½C after a long time

q = ½C as t→∞

6 0

3 years ago

Read 2 more answers

Estimate the average speed,in m/s,with which tge hair on your head grows. Make this estimate from your own experience nothing, f

Marat540 [252]

My estimate is 1/2 inch per week.

(0.5 inch/wk) x (1 wk / 7 da) x (1 da / 86,400 sec) x (1 meter / 39.37 inches)

= (0.5 x 1 x 1 x 1) / (7 x 86,400 x 39.37) meter/second

= 1 / 47,621,952 meter/second

= 0.0000000209987... m/s

= 2.1 x 10⁻⁸ meter/second

= 2.1 x 10⁻⁵ mm/sec

= 0.000021 millimeter per second (rounded)

7 0

3 years ago

Read 2 more answers

How do you solve this problem?

Katarina [22]

The particle has acceleration vector

$\vec a=\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\jmath$

We're told that it starts off at the origin, so that its position vector at $t=0$ is

$\vec r_0=\vec0$

and that it has an initial velocity of 12 m/s in the positive $x$ direction, or equivalently its initial velocity vector is

$\vec v_0=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\,\vec\imath$

To find the velocity vector for the particle at time $t$ , we integrate the acceleration vector:

$\vec v=\vec v_0+\displaystyle\int_0^t\vec a\,\mathrm d\tau$

$\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath$

$\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath$

Then we integrate this to find the position vector at time $t$ :

$\vec r=\vec r_0+\displaystyle\int_0^t\vec v\,\mathrm d\tau$

$\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath$

$\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath$