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rusak2 [61]
3 years ago
6

The noise floor, also known as additive white Gaussian noise (AWGN), is a continuous noise level that appears over a wide spectr

um when viewed in the frequency domain.
a. Trueb. False
Physics
1 answer:
harkovskaia [24]3 years ago
4 0

Answer:

correct option is a. True

Explanation:

solution

the noise floor is AWGN ( additive white Gaussian noise )  

and when viewed in the frequency domain, it is the continuous noise level  

because as they have a  uniform power over all the frequency.

 

so that it is additive white Gaussian noise  

as we can say given statement is True  

correct option a true  

You might be interested in
If you fall from a building onto a net which extends the time of impact by 10 times, what happens to the force you experience?
zubka84 [21]

Answer:

A larger impulse. A 1-kg ball has twice as much speed as a 10-kg ball.

Explanation:

6 0
3 years ago
What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a
Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

A = W + \dfrac{W}{2}

A =\dfrac{3W}{2}

A =\dfrac{3mg}{2}

When the car is at the bottom,  the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

N = m g + \dfrac{mv^2}{r}

\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}

\dfrac{mg}{2} = \dfrac{mv^2}{r}

v = \sqrt{\dfrac{rg}{2}}

v = \sqrt{\dfrac{40\times 9.8}{2}}

v = 14 m/s

8 0
3 years ago
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5.0 km/h
Ratling [72]

Answer

given,

time  = 10 s

ship's speed = 5 Km/h

F = m a

a is the acceleration and m is mass.

In the first case

F₁=m x a₁

where a₁ =  difference in velocity / time

F₁ is constant acceleration is also a constant.

Δv₁ = 5 x 0.278

Δv₁ = 1.39 m/s

a_1=\dfrac{1.39}{10}

a₁ = 0.139 m/s²

F₂ =m x a₂

F₃ = F₂ + F₁

Δv₃ = 19 x 0.278

Δv₃ = 5.282 m/s

a₃=Δv₂ / t

a_3=\dfrac{5.282}{10}

a₃ = 0.5282 m²/s

m a₃=m a₁ + m a₂

a₃ = a₂ + a₁

0.5282 = a₂ + 0.139

a₂=0.3892 m²/s

F₂ = m x 0.3892...........(1)

F₁ = m x 0.139...............(2)

F₂/F₁

ratio = \dfrac{0.3892}{0.139}

ratio = 2.8

6 0
3 years ago
In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.
Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

KE_i+PE_i =KE_f+PE_f

or

\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

g=9.8 m/s^2 is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s

4 0
3 years ago
Convert the speed of light 3.0x10^8 m/s to km/day
Aleksandr [31]

Answer: 2.592 \times 10^{8}km/day

1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day



7 0
3 years ago
Read 2 more answers
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