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3 years ago

Choose the statement that is a misconception about a system in chemical equilibrium.

Physics

1 answer:

Maksim231197 [3]3 years ago

4 0

Answer:

There are equal parts reactants and products.

Explanation:

When a system is in a chemical equilibrium, the rate at which the forward reaction occur is the same as the rate at which the reverse reaction occurs. This system is said to undergo a reversible reaction. The concentration of the reactants and products do not change, i.e are constant.

In a chemical equilibrium, the reaction continues in both directions but at the same rate.

Therefore the statement that is a misconception about a system in chemical equilibrium is 'There are equal parts reactants and products'.

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Plan an experiment to measure the ideal mechanical advantage of a three-hole punch. (a) What materials would you need? (b) What

Vaselesa [24]

Answer:

A) Three hole punch and either a layered plastic or paper

B) Identify the lengths involved ,

Length of input arm / length of output arm = L1/ L2

Explanation:

a) Materials involved includes :

Three hole punch and either a layered plastic or paper

Identify the forces acting on the three-hole punch which are Input and output forces

Identify the points where they act

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3 years ago

A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C an

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Answer:

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV................ Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

E = CV²/2............. Equation 2

Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

E = 4 J

Hence the right option is c. 0.02 C and 4 J

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2 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago