All categories

3 years ago

What type of noise is most difficult to remove from a digital signal? why?

Physics

1 answer:

Lilit [14]3 years ago

8 0

Interference if the hardest to remove due to the superposition of the frequencies of the radio waves

You might be interested in

An object with a mass of 6.0 kg accelerates 8.0 m/s^2 when an unknown force is applied to it. What is the amount of the force? R

iren [92.7K]

Answer:

48N

Explanation:

use F=ma, or force is equal to mass multiplied by acceleration.

8 0

3 years ago

Which renewable resource produces the most electricity in the United States? geothermal energy solar energy wind power hydroelec

Anna71 [15]

Hydroelectricity is the best answer.

This is an article by the EIA, but the pie graph is the most helpful: https://www.eia.gov/energyexplained/?page=us_energy_home

8 0

3 years ago

Read 2 more answers

In the diagram of the conical pendulum below, which number (1, 2, 3, or 4) represents the angle θ used in the conical pendulum e

Mama L [17]

Step 1: Look in your book or online for the conical pendulum equation.

Step 2: Look at the drawing and see which angle is involved in the equation.

Answer: It's Angle #2 in your drawing.

8 0

3 years ago

Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If

jeka57 [31]

Let $F_{1}=F_{2}=F$ .

Normal force equals (using Newton's third law) $N=mg+F\sin30^{o}-F\sin37^{o}$ .

$F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o})$ , but $F_{f}\leq F(\cos30^o+\cos37^o)$ for all $F_{f}$ (in order to start moving the break). Therefore $F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o})$ , solving for $F$ : $F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}$

7 0

3 years ago

An object moves uniformly around a circular path of radius 19.0 cm, making one complete revolution every 2.40 s.

klio [65]

Answer:

v = 0.5 m/s

f = 0.42 Hz

ω = 2.6 rad/sec

Explanation:

By definition, the translational speed is the rate of change of the position with respect to time.
The change in position along a complete revolution is just the following:
Δs = 2*π*r = 2*π*0.19 m = 1.19 m
The time needed to complete a revolution is 2.4 s, so the translational speed can be written as follows:

$v =\frac{\Delta s}{\Delta t} = \frac{1.19m}{2.4s} = 0.5 m/s (1)$

The frequency in Hz is just the inverse of the time needed to complete a revolution (known as the period T), as follows:
f = 1/T = 1/2.4s = 0.42 Hz (2)
Finally, the angular speed is the rate of change of the angle rotated with respect to time, as follows:

$\omega = \frac{\Delta\theta}{\Delta t} = \frac{2*\pi}{2.4s} = 2.6 rad/sec (3)$

5 0

3 years ago