I believe the answer is A.
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:
The position is relative to the base of the ocean
Explanation:
From the question we are told that
The angle made by the incline with the horizontal is
The constant acceleration is
The distance covered is
The height of the cliff is
The velocity of the car is mathematically represented as
The initial velocity of the car is u= 0
So
substituting values
The vertical component of this velocity is
substituting values
The negative sign is because is moving in the negative direction of the y-axis
The horizontal component of this velocity is
Now according to equation of motion we have
substituting values
using quadratic equation we have that
given that time cannot be negative
The car’s position relative to the base of the cliff when the car lands in the ocean is mathematically evaluate as
substituting values
The component of weight(
mg) that is responsible for the motion of boxes on ramp is:
Where m = mass of the boxes.
g = Acceleration due to gravity = 9.8
= The angle the ramp makes with the ground. In this case it is 30°.
Since the frictional force is:
μ*N.
Where,
μ = Frictional Coefficient
N = Normal to the ramp =
Therefore, the frictional force becomes =
= μ*
Apply Newton's second law we would get:
- μ*
=
=>
=
μ
-- (A)
Now according to equation of motion:
Where x = 5.8m
= 0
= 0
= 10.24
Plug in the value in the above equation you would get:
= 1.1328
Plug in
in equation (A) and solve for
μ, you would get,
μ = 0.4439
Answer:
The greater the luminosity of a star, the longer its period of oscillation.