Answer:
9.81 × 10⁻¹⁰ C
Explanation:
Given:
Distance between the tissue and the tip of the scale, r = 6 cm = 0.06 m
Charge on the ruler, Q = - 12 μC = - 12 × 10⁻⁶ C
Mass of the tissue = 3 g = 0.003 Kg
Now,
The force required to pick the tissue, F = mg
where, g is the acceleration due to gravity
also,
The force between (F) the charges is given as:

where,
q is the charge on the tissue
k is the Coulomb's constant = 9 × 10⁹ Nm²/C²
thus,

on substituting the respective values, we get

or
q = 9.81 × 10⁻¹⁰ C
Minimum charge required to pick the tissue paper is 9.81 × 10⁻¹⁰ C
Answer:
2.72 cycles
Explanation:
First of all, let's find the time that the stone takes to reaches the ground. The stone moves by uniform accelerated motion with constant acceleration g=9.8 m/s^2, and it covers a distance of S=44.1 m, so the time taken is

The period of the pendulum instead is given by:

Therefore, the number of oscillations that the pendulum goes through before the stone hits the ground is given by the time the stone takes to hit the ground divided by the period of the pendulum:

so you can see all the different categories at once. both as a whole and on an individual scale.
A spinning wheel
and the blade of a kitchen blender both illustrate rotation.
A child swinging on a swing
illustrates oscillation, or 'harmonic' motion.
A balloon being blown up is an example of dilation or inflation.
A sliding hockey puck demonstrates the concept of translation.
Answers:(a) 
μT
(b) 
μm
(c) f =
Explanation:Given electric field(in y direction) equation:

(a) The amplitude of electric field is

. Hence
The amplitude of magnetic field oscillations is

Where c = speed of light
Therefore,

μT (Where T is in seconds--signifies the oscillations)
(b) To find the wavelength use:



μm
(c) Since c = fλ
=> f = c/λ
Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f =