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olya-2409 [2.1K]

3 years ago

14

What effect does a catalyst have on a system in equilibrium?

1 answer:

Jet001 [13]3 years ago

8 0

ANSWER:

What effect does a catalyst have on a system in equilibrium?

The system is unaffected.

~batmans wife dun dun dun....

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a positively charged body makes contact with a body. after a while, the charged body becomes neutralised. state 3 main condition

MrRa [10]

Answer:

A body will become positively charged when some electrons will come out from the body.Thus, positive charge is due to deficiency of electrons.

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2 years ago

1.20 Newton force is working on a 250 gram object. What is the acceleration?

Leokris [45]

Answer:

The answer is B

Explanation:

250g = 0.25kg

F = m × a

a = F/m

= 1.2/0.25

= 4.8m/s²

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2 years ago

Does centripetal force change in uniform circular motion?

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It depends on how fast you are going and in orincipal no

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3 years ago

Which elements do hydrogen fuel cells combine to produce electricty

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2 years ago

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

3 years ago