Answer: distance d = 4.73e10m
Explanation: Suppose the charge on the black hole is 5740 C which is a positive charge.
Using electric potential V formula:
V = kq / d
Where K = 9.05×10^9Nm^2/C
And e = 1.6×10^-19C
But you don't need to substitute it.
1090 V = 8.99e9N·m²/C² * 5740C /d
Make d the subject of formula
d = 4.73e10 m
(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in
(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s
(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:
(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s
(c) The plane accelerates toward the center of the path with magnitude
<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²
(d) By Newton's second law, the tension in the line is
<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N
We are given:
v0 = initial velocity = 18 km/h
d = distance = 4 km
v = final velocity = 75 km/h
a =?
<span>
We can solve this problem by using the formula:</span>
v^2 = v0^2 + 2 a d
75^2 = 18^2 + 2 (a) * 4
5625 = 324 + 8a
<span>a = 662.625 km/h^2</span>
Biology, physics, geology
Complete Question:
When specially prepared Hydrogen atoms with their electrons in the 6f state are placed into a strong uniform magnetic field, the degenerate energy levels split into several levels. This is the so called normal Zeeman effect.
Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 Tesla?
Answer:
ΔE = 1.224 * 10⁻²² J
Explanation:
In the 6f state, the orbital quantum number, L = 3
The magnetic quantum number, 
The change in energy due to Zeeman effect is given by:
Magnetic field B = 2.02 T
Bohr magnetron, 
ΔE = 1.224 * 10⁻²² J
