Answer:
1 million hahahahahahahahhahah
Divide
(the distance covered in some period of time)
by
(the time taken to cover the distance).
The quotient is the average speed during that period of time.
Answer:
Micro and radio waves.
Lower energy.
Gamma rays.
Explanation:
The electromagnetic spectrum is the range of frequencies of electromagnetic radiation and their respective wavelengths.
Ionising radiation os defined as the energy required of photons of a wave to ionize atoms, causing chemical reactions.
The energy of the wave depends on both the amplitude and the frequency. If the energy of each wavelength is a discrete packet of energy, a high-frequency wave will deliver more of these packets per unit time than a low-frequency wave. In summary, the longer the wavelength, the lower the energy to ionise.
The velocity of a wave is directly proportional to the frequency of that wave.
c = f * lambda
Where,
c = velocity of the wave
f = frequency of the wave = 1/time
Lambda = wavelength.
From the above expression, the longer the wavelength, lambda the shorter the frequency.
Examples of waves with longer wavelengths are, micro and radio waves, while radiations with shorter wavelengths like gamma rays.
Answer:
The acceleration is 1 cm/s^2.
Explanation:
The acceleration is defined as the rate of change of velocity.
Here, initial velocity, u = 3/1 = 3 cm/s
final velocity, v = 4/1 = 4 cm/s
time, t = 1 s
Let the acceleration is a.
Use first equation of motion
v = u + at
4 = 3 + 1 x a
a = 1 cm/s^2
The equation of motion of a pendulum is:
where 
it its length and 
is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles (
), we can use:
Additionally, let us define:
We can now write:
The solution to this differential equation is:
where 
and 
are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:
This justifies that the period depends only on the pendulum's length.
