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3 years ago

Select the correct answer.

Physics

1 answer:

Natali5045456 [20]3 years ago

6 0

(NH4)2CO3(s) ⇄ 2NH3(g) + CO2(g) + H2O (g)

The effect that adding water (H2O) have in reaction above is that The concentration of the product would decrease, and the concentration of reactant would increase. (answer C)

<u><em>Explanation</em></u>

If the concentration of one substance is changed, the equilibrium shift to minimize the effect of that change.

H2O in reaction above is a product.

Increase of water (H2O) concentration will led to the shift of equilibrium to the left ( reactant side) .The concentration of reactant increase as that of product decrease.

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Where do most metamorphic processes take place?

inysia [295]

Answer:

Most metamorphic processes take place deep underground, inside the earth's crust.

Explanation:

During metamorphism, protolith chemistry is mildly changed by increased temperature (heat), a type of pressure called confining pressure, and/or chemically reactive fluids. hope this helps you :)

8 0

3 years ago

A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate

yaroslaw [1]

Answer:

The maximum error is $\Delta \eta = 2032.9$

Explanation:

From the question we are told that

The length is $l = 1\ m$

The radius is $r = 0.002 \pm 0.0002 \ m$

The pressure is $P = 4 *10^{5} \ \pm 1750$

The rate is $v = 0.5*10^{-9} \ m^3 /t$

The viscosity is $\eta = \frac{\pi}{8} * \frac{P * r^4}{v}$

The error in the viscosity is mathematically represented as

$\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v$

Where $\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}$

and $\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}$

and $\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}$

$\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]$

substituting values

$\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]$

$\Delta \eta = \frac{\pi}{8} [56 + 5120 ]$

$\Delta \eta = 647 \pi$

$\Delta \eta = 2032.9$

4 0

3 years ago

Write a speech about bringing back fidget toys because they're banned in school

12345 [234]

I believe that fidget toys should not be brought back in school, reason being is because fidget spinners were originally made for students with example: Autism . Who actually needed them. Now that people who don't need them are buying them and playing with them in the classroom, it causes a complete distraction and is not being used for its original purpose whatsoever .

7 0

3 years ago

There are two blocks: one large one initially at rest, and a smaller one, initially moving to the right withsome speed. The smal

KATRIN_1 [288]

Answer:

Explanation:

Let the initial velocity of small block be v .

by applying conservation of momentum we can find velocity of common mass

25 v = 75 V , V is velocity of common mass after collision.

V = v / 3

For reaching the height we shall apply conservation of mechanical energy

1/2 m v² = mgh

1/2 x 75 x V² = 75 x g x 10

V² = 2g x 10

v² / 9 = 2 x 9.8 x 10

v² = 9 x 2 x 9.8 x 10

v = 42 m /s

small block must have velocity of 42 m /s .

Impulse by small block on large block

= change in momentum of large block

= 75 x V

= 75 x 42 / 3

= 1050 Ns.

6 0

3 years ago

Two male moose charge at each other with the same speed and meet on a icy patch of tundra. as they collide, their antlers lock t

yanalaym [24]

Their linear inertia is equivalent to their masses. Let the inertia of the first moose be m₁ and the second be m₂.

m₁u + m₂u = (m₁ + m₂) x 1/3 u
3m₁ + 3m₂ = m₁ + m₂
3 m₁/m₂ + 3 = m₁/m₂ + 1
m₁/m₂ = 2

The ratio of their inertias is 2

7 0

3 years ago