Answer:
20. This is a transverse wave.
21. a is a Crest
b is the wavelength
c is the amplitude
d is the trough
e isthe amplitude
f is the wavelength
g is time
Explanation:
The period of a simple pendulum is given by
![T= 2 \pi \sqrt{ \frac{L}{g} }](https://tex.z-dn.net/?f=T%3D%202%20%5Cpi%20%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg%7D%20%7D%20)
where
L is the pendulum length
g is the acceleration of gravity
If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
![T_e= 2 \pi \sqrt{ \frac{L}{g_e} }](https://tex.z-dn.net/?f=T_e%3D%202%20%5Cpi%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg_e%7D%20%7D%20)
where
![g_e](https://tex.z-dn.net/?f=g_e)
is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
![T_m= 2 \pi \sqrt{ \frac{L}{g_m} }](https://tex.z-dn.net/?f=T_m%3D%202%20%5Cpi%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg_m%7D%20%7D%20)
where
![g_m](https://tex.z-dn.net/?f=g_m)
is the acceleration of gravity on the Moon.
If we do the ratio of the two periods, we get
![\frac{T_m}{T_e} = \sqrt{ \frac{g_e}{g_m} }](https://tex.z-dn.net/?f=%20%5Cfrac%7BT_m%7D%7BT_e%7D%20%3D%20%20%5Csqrt%7B%20%5Cfrac%7Bg_e%7D%7Bg_m%7D%20%7D%20%20)
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write
![g_e = 6 g_m](https://tex.z-dn.net/?f=g_e%20%3D%206%20g_m)
and we can rewrite the previous ratio as
![\frac{T_m}{T_e} = \sqrt{ \frac{6 g_m}{g_m} }= \sqrt{6}](https://tex.z-dn.net/?f=%5Cfrac%7BT_m%7D%7BT_e%7D%20%3D%20%5Csqrt%7B%20%5Cfrac%7B6%20g_m%7D%7Bg_m%7D%20%7D%3D%20%20%5Csqrt%7B6%7D%20)
so the period of the pendulum on the Moon is
Answer:
your questions are not clear
Answer:
C. difference between the highest and lowest frequencies that can be accommodated on a single channel.
Explanation:
Bandwidth is the range of the band of frequencies. It is the amount of the data which can be transmitted over a wide range of frequency in the fixed time.
Bandwidth is difference between highest and lowest frequencies in the continuous band of the frequencies which can be accommodated on the single channel. It is measured in hertz.
<u>Correct answer - C. difference between the highest and lowest frequencies that can be accommodated on a single channel.</u>
Answer:
increasing the masses of the objects and decreasing the distance between them
Explanation:
F = GmM/d²
Their masses are in the numerator, increased mass increases the force
Their separation is in the denominator, shrinking distance increases the force.