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3 years ago

The law of mass action suggests that _____.

Physics

2 answers:

Nostrana [21]3 years ago

7 0

the higher concentration of molecules, the faster a reaction can occur

pochemuha3 years ago

6 0

Answer:

<h2>The higher concentration of molecules, the faster a reaction can occur</h2>

Explanation:

The law of mass action states that the rate of a chemical reaction is directly proportional to the product of the masses of the reactor. In other words, the more molecules concentration, the faster the reaction can develop.

Therefore, the choice that completes the given statement is the first one: "the higher concentration of molecules, the faster a reaction can occur".

So, the completed statement is<em> The law of mass action suggests thatthe higher concentration of molecules, the faster a reaction can occur.</em>

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A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

3 years ago

Which of the following examples illustrates static friction?

vivado [14]

Answer:

A box sits stationary on a ramp

Explanation:

Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.

Static force of friction is calculated as follows:

F= μη

F is static force of friction.

μ is the coefficient of static friction.

η is the normal force.

6 0

4 years ago

Is light fastest than speed and why.

Genrish500 [490]

Answer:

Explanation:

There is no limit to how fast the universe can expand, says physicist Charles Bennett of Johns Hopkins University. Einstein's theory that nothing can travel faster than the speed of light in a vacuum still holds true, because space itself is stretching, and space is nothing.

7 0

3 years ago

The air pressure at the ocean surface a few miles from the shoreis most likely

frez [133]

The air pressure is most likely lower.

8 0

3 years ago

Ron is on a Ferris wheel of radius 30 ft that turns counterclockwise at a rate of one revolution every 12 seconds. The lowest po

alexgriva [62]

Answer:

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

Explanation:

1 full revolution is $2\pi.$ let \theta be the angle of Ron's position.

At t = 0. $\theta = 0$

one full revolution occurs in 12 sec, so his angle at t time is

$\theta =2\pi \frac{t}{12} = \frac{\pi}{6}t$

r is radius of circle and it is given as

$x = rcos\theta$

$y = rsin\theta$

for r = 30 sec

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t$

however, that is centered at (0,0) and the positioned at time t = 0 is (30,0). it is need to shift so that the start position is (30,45). it can be done by adding to y

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

7 0

3 years ago