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inn [45]
3 years ago
10

In the presence of ________, hydrogen peroxide will decompose into oxygen and water.

Physics
1 answer:
IrinaK [193]3 years ago
7 0

Answer: Catalase enzyme

Explanation:

The catalase is an enzyme that is commonly found in the living organisms that are exposed to minute or high amount of oxygen or those in which aerobic respiration takes place.

This enzyme catalyzes the breakdown or decomposition of the hydrogen peroxide into oxygen and water. In humans and animals the catalase is present in liver, where it conducts the function of detoxifying the body by decomposing the toxins.

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Please help with Physics Circuits!
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1) Let's start by calculating the equivalent resistance of the three resistors in parallel, R_2, R_3, R_4:
\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}
From which we find
R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega
So, the correct answer is D) 


2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} =  \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}
From which we find
R_{23} = 2.5 \Omega

And these are connected in series with a resistor of 10 \Omega, so the equivalent resistance of the circuit is
R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega

And by using Ohm's law we find the current in the circuit:
I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A
So, the correct answer is C).


3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} =  \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}
From which we find
R_{23} = 2.5 \Omega
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega

And by using Ohm's law we find the current flowing in the circuit:
I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V
So, the correct answer is D).


4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}
From which we find
R_{23} = 7.55 \Omega

Now all the resistors are in series, so the equivalent resistance of the circuit is:
R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega

The current in the circuit is given by Ohm's law
I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A

Now we can compare the voltage drops across the resistors. Resistor 1:
V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V
Resistor 4:
V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V

So, the greatest voltage drop is on resistor 1, so the correct answer is D).


5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
P_3 = I_3^2 R_3 =  \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.
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