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Lelu [443]
3 years ago
11

Which statement describes a digital signal?

Physics
2 answers:
max2010maxim [7]3 years ago
5 0

Answer:it is made of number

Explanation:

I got it right

hoa [83]3 years ago
4 0

Answer:

"It is made of numbers" describes the digital signal.

Explanation:

The digital signal are the electrical signal which is translated into the pattern of bits. The digital signal are always discrete value in every sampling point. The conversion of the programming into the stream or the binary sequence like 0s and 1s. The digital signals never gets weaken over distance but the analog signal gets weakened or impair at distance. The digital signals are consists of one or two value, Timing graph are square waves.

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5. A person fishing from a pier observes that four wave crests pass by in 7.0 s and estimates that the distance between two succ
TiliK225 [7]

Answer:

v= 1.71 m/s

Explanation:

Given that

Distance between two successive crests = 4.0 m

 λ = 4 m

T= 7 sec

T is the time between 3 waves.

3 waves = 7 sec

1 wave = 7 /3 sec

So t= 7/3 s

We know that frequency f

f= 1/t= 3/7 Hz

Lets take speed of the wave is v

v= f λ

f=frequency

λ=wavelength

v= 3/7 x 4 = 12 /7

v= 1.71 m/s

3 0
3 years ago
A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful
balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T

the angle between B and L is given by:

\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N

hence, the magnitude of the magnetic force is 0.187N

4 0
3 years ago
Which of the following is a metal? Calcium (Ca) Iron (Fe) Sodium (Na) all of these
ANTONII [103]
Iron (Fe) or at least it can be.

8 0
3 years ago
Read 2 more answers
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
A block and tackle is used to lift an automobile engine that weighs 1800 N. The person exerts a force of 300 N to lift the engin
Sidana [21]

Answer:

1800/300 = 6ropes

Explanation:

The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.

8 0
3 years ago
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