Ask question

Ask question

All categories

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....

You might be interested in

What was the major shortcoming of rutherford’s model of the atom?

lisabon 2012 [21]

The major shortcoming of Rutherford's model was that it was incomplete. It did not explain how the atom's negatively charged electrons are distrubuted in the space surronding its positively charged nucleus. A form of energy that exhibits wavelike behavior as it travels through space

7 0

2 years ago

Given a 10-V power supply, would a 20-ohm resistor and a 5-ohm resistor need to be arranged in parallel or in series to generate

Bogdan [553]

Answer:

The resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

Explanation:

We are given 10 voltage power source and we have two Resistors with resistance of 20 ohm and 5ohm.

We need to find the orientation in which these two resistors would be arranged so that the circuit could get a current of 2.5Ampere.

Using ohm's law we have

V = I*R

V= voltage

I= current

R= resistance

10 = 2.5*R

R = 10/2.5 = 4ohm

that means we need a total of 4ohm resistance from these two resistors.

since the net Resistance(4ohm) is lower than the smallest resistance(5ohm) available that means the orientation of the resistors will be in parallel.

$\frac{1}{R} = \frac{1}{R1} +\frac{1}{R2}\\ = \frac{1}{20} +\frac{1}{5}\\ \\ = \frac{1+4}{20} =\frac{1}{4}$

R(net) =4ohm

Now the orientation of the resistors are in parallel so the current will be divided.

we know that the current will divide in opposite manner the arm which provides more resistance less current will flow from there and vice versa.

We know that the voltage in parallel remains same

In 20 ohm resistance

again using ohms law

V = i1*R1

10 = i1*20

i1 = 0.5A

in 5ohm resistor

V=i2*R2

10 = I2*5

i2 =2A

and i1+i2 = 0.5+2= 2.5A which means our calculation is correct.

Therefore the resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

6 0

2 years ago

Unscramble the bolded letters to guess the 6 letter word code. Did you get the poses or exercises correct? MOUNTAIN POSE TRIANGL

nikitadnepr [17]

Explanation:

I don't see a question...

8 0

2 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago

Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th

Ket [755]

Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

5 0

3 years ago