Answer:
There will be two forces acting on her: Gravitational force and Air resisitence
Answer:
1. The period is 1.74 s.
2. The frequency is 0.57 Hz
Explanation:
1. Determination of the the period.
Spring constant (K) = 30 N/m
Mass (m) = 2.3 Kg
Pi (π) = 3.14
Period (T) =?
The period of the vibration can be obtained as follow:
T = 2π√(m/K)
T = 2 × 3.14 × √(2.3 / 30)
T = 6.28 × √(2.3 / 30)
T = 1.74 s
Thus, the period of the vibration is 1.74 s.
2. Determination of the frequency.
Period (T) = 1.74 s
Frequency (f) =?
The frequency of the vibration can be obtained as follow:
f = 1/T
f = 1/1.74
f = 0.57 Hz
Thus, the frequency of the vibration is 0.57 Hz
Answer:
(a) False
(b) True
(c) True
(d) True
(e) True
(f) True
Explanation:
(a) Maxwell's equations not only applies to constant fields but it applies to both the fields, i.e., Time variant field as well as Time Invariant field.
(b) We make use of the Modified form of the Ampere's law and Faraday's Law to derive the wave equation.
(c) Electromagnetic waves contains both the electric and magnetic fields and these fields oscillates at an angle of
to the direction of wave propagation.
(d) In free space both the electric and magnetic fields are in phase while considering electromagnetic waves.
(e) In free space or vacuum, the expression for the speed of light in terms of electric and magnetic field is given as:

Thus the ratio of the magnitudes of the electric and magnetic field vectors are equal to the speed of light in free space.
(f) In free space or in vacuum the energy density of the electromagnetic wave is divided equally in both the fields and hence are equal.
Answer:

Explanation:
To convert the energy of the cereal to know the weight:






Supuse the gravity as a g=9.8m/s^2


Answer:
The point at which the electrical potential is zero is x = +0.33 m.
Explanation:
By definition the electrical potential is:

Where:
K: is Coulomb's constant = 9x10⁹ N*m²/C²
q: is the charge
r: is the distance
The point at which the electrical potential is zero can be calculated as follows:

(1)
q₁ is the first charge = +3 mC
r₁ is the distance from the point to the first charge
q₂ is the first charge = -6 mC
r₂ is the distance from the point to the second charge
By replacing r₁ = 1 - r₂ into equation (1) we have:
(2)
By solving equation (2) for r₂:

Therefore, the point at which the electrical potential is zero is x = +0.33 m.
I hope it helps you!