Answer:
The most precise is 0.0034 m ⇒ 1st answer
Explanation:
<em>Lets revise the meaning of precise</em>
Precision is referred as the closeness of two measurements to one
another.
Highest number with significant digit is the most precise
Ex: <em>Highest significant </em>of optical instrument is the Wavelength of visible (red)
light = 0.00006 cm
Hence the <em><u>most precise</u></em> device for measuring length is the given optical
instrument
The difference between accuracy and precise
<em>Accuracy</em> is how close a value is to its true value
<em>Precision</em> is how repeatable a measurement is
We have three numbers
0.0034 m, 45.6 m, and 1234 m
The least precise is 1234 m
The most precise is 0.0034 m
To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:
= Coulomb's constant
q = Charge
r = Radius
At the same time
The values of variables are the same, then if we replace in a single equation we have this expression,
If we replace the values, we have finally that the charge is,
Therefore the potential energy of the system is 
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
It’s North of the equator
