The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
<h3>
Total capacitance of the circuit</h3>
The total capacitance of the circuit is calculated as follows;
Capacitors in series;
1/Ct = 1/8 + 1/7.5
1/Ct = 0.25833
Ct = 3.87 mF
Capacitors is parallel;
Ct = 3.87 mF + 12 mF + 15 mF
Ct = 30.87 mF
Ct = 0.03087 F
<h3>Charge in each capacitor</h3>
Q = CV
Q = 0.03087 x 12
Q = 0.37 C
Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
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Answer:
Answer:
Q_1 = 7Q
1
=7
Q_2 = 10Q
2
=10
Q_3 = 13.5Q
3
=13.5
Step-by-step explanation:
Given
5, 7, 7, 8, 10, 11, 12, 15, 17.
Required
Determine Q1, Q2 and Q3
The number of data is 9
Calculating Q1:
Q1 is calculated as:
Q_1 = \frac{1}{4}(N + 1)Q
1
=
4
1
(N+1)
Substitute 9 for N
Q_1 = \frac{1}{4}(9 + 1)Q
1
=
4
1
(9+1)
Q_1 = \frac{1}{4}*10Q
1
=
4
1
∗10
Q_1 = 2.5th\ itemQ
1
=2.5th item
This means that the Q1 is the mean of the 2nd and 3rd data.
So:
Q_1 = \frac{1}{2}(7+7)Q
1
=
2
1
(7+7)
Q_1 = \frac{1}{2}*14Q
1
=
2
1
∗14
Q_1 = 7Q
1
=7
Calculating Q2:
Q2 is calculated as:
Q_2 = \frac{1}{2}(N + 1)Q
2
=
2
1
(N+1)
Substitute 9 for N
Q_2 = \frac{1}{2}(9 + 1)Q
2
=
2
1
(9+1)
Q_2 = \frac{1}{2}*10Q
2
=
2
1
∗10
Q_2 = 5th\ itemQ
2
=5th item
Q_2 = 10Q
2
=10
Calculating Q3:
Q3 is calculated as:
Q_3 = \frac{3}{4}(N + 1)Q
3
=
4
3
(N+1)
Substitute 9 for N
Q_3 = \frac{3}{4}(9 + 1)Q
3
=
4
3
(9+1)
Q_3 = \frac{3}{4}*10Q
3
=
4
3
∗10
Q_3 = 7.5th\ itemQ
3
=7.5th item
This means that the Q3 is the mean of the 7th and 8th data.
So:
Q_3 = \frac{1}{2}(12+15)Q
3
=
2
1
(12+15)
Q_3 = \frac{1}{2}*27Q
3
=
2
1
∗27
Q_3 = 13.5Q
3
=13.5
Answer:
the answer is A because I did this in 5th grade
The GPE is converted into kinetic energy as it falls.
Answer:
The relative density of the second liquid is 7.
Explanation:
From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.
Let us assume that the volume of the object is 'V'
Thus for the liquid in which the block is completely submerged
The buoyant force should be equal to weight of liquid
Mathematically
Thus for the liquid in which the block is 1/7 submerged
The buoyant force should be equal to weight of liquid
Mathematically
Comparing equation 'i' and 'ii' we see that
Since the first liquid is water thus 
Thus the relative density of the second liquid is 7.
