w = mg where g = 9.8 N/kg
m = 580/9.8 = 59 kg
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
4.266 m is the radius of the circular path the electron follows.
Given
Speed of electron (v) = 7.5 × 10⁶ m/s
Earth's Magnetic Field (B) = 1 × 10⁻⁵ T
We already know that
Mass of electron (m) = 9.1 × 10⁻³¹ kg
Charge on electron (q) = 1.6 × 10⁻¹⁹ C
According to the formula
Radius of circular path(r) = mass on electron × speed/ Charge × Magnetic field
Radius of circular path(r) = m × v/q × B
Put the values into the formula
r = 9.1 × 10⁻³¹ × 7.5 × 10⁶/ 1.6 × 10⁻¹⁹ × 10⁻⁵
On solving, we get
r = 4.266 m
Hence, 4.266 m is the radius of the circular path the electron follows.
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