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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

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

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The amount of WORK done is determined by 2 factors.

RUDIKE [14]

Explanation :

Work is done when a force is applied to create a displacement on an object.

Thus, the work done depends on the two factors i.e.

(1) Applied force (F)

(2) Distance or displacement (d)

Mathematically, work done is $W= F.s$

It also depends on the angle between the force and the displacement.

$W=Fs\ cos\ \theta$

For example,

A person carries a weight of 20 kg and lifts it on his head 1.5 m above the surface. So, the work done by him on the luggage will be:

$W= F\times s$

or

$W=m\times g \times s$

$W=20\ kg\times 9.8\ m/s^2\times 1.5\ m$

So, $W=294\ Joules$

Hence, the work done by him on the luggage is 294 Joules.

6 0

3 years ago

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

$h=v_yt-\frac{1}{2}gt^2$

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0

3 years ago

We decided to make an iced latte by adding ice to a 200 mL hot latte at 45 °C. The ice starts out at 0 C. How much ice do we nee

tankabanditka [31]

Answer:

m = 77.75 g

Explanation:

Here we know that at equilibrium the temperature of the system will be 10 degree C

so heat given by hot latte = heat absorbed by the ice

now we have

heat given by latte = $m s\Delta T$

$Q_1 = (200)(4.186)(45 - 10)$

$Q_1 = 29302 J$

now heat absorbed by ice is given as

$Q_2 = mL + ms\Delta T$

$Q_2 = m(335 + 4.186(10 - 0))$

$Q_2 = m(376.86)$

now by heat balance we have

$Q_1 = Q_2$

$29302 = m(376.86)$

$m = 77.75 g$

6 0

2 years ago

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

So

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

2)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

So

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

2 years ago