What is one disadvantage of sending information over long distances
wirelessly using digital signals?
O A. The signals become weaker the farther the receiver is from the
source.
B. The farther the signals travel, the more slowly they move.
C. The signals become stronger the farther the receiver is from the
source.
O D. The farther the signals travel, the easier they are to detect.

Answers

Answer 1

The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."

Since most of the signal which we use for communication are radio signal Radio signal are basically electromagnetic waves.As the wave moves forward it looses its amplitude.So basically radio signal becomes weaker for long distance.

What are radio signals?Radio signals or radio waves are a form of electromagnetic wave. Although this may sound complicated, it is possibly sufficient to say that these waves have both electric and magnetic components. They are the same as light rays, ultra-violet and infra-red. The only difference is in the wavelength of the waves.

Thus , The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."

Learn more about Radio waves here -

https://brainly.com/question/69373

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Related Questions

Fossil clues are one of the _____________ clues that support the theory of continental drift.
A. crust B. resource C. climate D. rock

Answers

Answer:

a

Explanation:

I think don't get mad if I'm wrong

PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE



Which best explains why species living in Australia are found nowhere else on Earth? This is an example of Geologic Evolution.

A.
Australia has an ecosystem different from any other area on Earth.

B.
Humans have genetically altered many Australian species in laboratories.

C.
Australian species were genetically altered after a comet hit the landmass.

D.
Australia separated from other continents and species there evolved independently.

Answers

Australia separated from other continents and species there evolved independently

A scientist analyzes the light from a distant galaxy and finds that it is shifted to the longer wavelength of the electromagnetic spectrum. What does this data help to study?
1) the color of the galaxy
2) the distance of the galaxy from Earth
3) the existence of life on any planet in the galaxy
4) the study of the amount of light scattered by dust in space

Answers

Answer:

Option 2

Explanation:

As per the relation between the distance of the galaxy and shifting of the light of the galaxy towards any specific wavelength of the electromagnetic spectrum, a galaxy at great distance shifts more towards the red spectra that has the highest wavelength.

Thus, this observation give details about the distance of the galaxy from earth.

Answer:

b

Explanation:

The Sun is divided into three regions.
True оr False?

Answers

Answer:

false I think

Explanation:

hope that help

so it's not divided in 3 regions

HURRYYYYY, FASTTT, HELPPP What is the consequence of atoms being a singular closed system?


Atoms of a certain element contain a fixed amount of energy.


Atoms of a certain element can’t be altered by outside forces.


Atoms don’t interact with each other in any way.


Atoms decompose slowly over time as they lose energy.


1

Answers

Answer:

Atoms of a certain element contain a fixed amount of energy.

Explanation:

An atom can be defined as the smallest unit comprising of matter that forms all chemical elements. Thus, atoms are basically the building blocks of matters and as such defines the structure of a chemical element.

Generally, these atoms are typically made up of three distinct particles and these are protons, neutrons and electrons.

Additionally, all the physical properties of a mineral result from the mineral's internal arrangement of atoms.

The consequence of atoms being a singular closed system is that atoms of a certain element contain a fixed amount of energy because the number of protons, neutrons and electrons are in a fixed proportion.

Let's assume raspberries are 10 wt% protein solids and the remainder water. When making jam, raspberries are crushed and mixed with sugar, in a 45:55 berry to sugar ratio, by mass. Afterward, the mixture is heated, boiling off water until the remaining mixture is 0.4 weight fraction water, resulting in the final product, jam. How much water, in kilograms, is boiled off per kilogram of raspberries processed

Answers

Answer:

The mass of water boiled off is [tex]0.0 \overline{185}[/tex] kg

Explanation:

The given percentage by weight of protein solids in raspberries = 10 weight%  

The ratio of sugar to raspberries in ja-m = 45:55

The mass of the mixture after boiling = 0.4 weight fraction water

Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry

The mass of raspberry, r = 1 kg

The percentage by weight of water in raspberry = 90 weight %

The mass of water in 1 kg of raspberry =  90/100 × 1 kg = 0.9 kg

The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55

∴ s = 1 kg × 55/45 = 11/9 kg

The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg

The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction

Let 'w' represent the mass of water boiled off, we have;

(0.9 - w)/(20/9 - w) = 0.4

(0.9 - w) = 0.4 × (20/9 - w)

0.9 - w = 8/9 - 0.4·w

9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w

(81 - 80)/(90) = (6/10)·w

1/90 = (6/10)·w

w = ((10/6) × 1/90) = 1/54

w = 1/54  

The mass of water boiled off, w = (1/54) kg = [tex]0.0 \overline{185}[/tex] kg

Build a second circuit with a battery and a light bulb but this time add a switch. Your circuit might look something like the one at right. When a switch is open in a circuit, it means the two ends are disconnected and current cannot flow between them. When a switch is closed in a circuit, it means the two ends are connected and current can flow between them. Play with the switch to check how it affects the flow of current. With the switch closed, compare the brightness of the bulb and the flow of current in this circuit, with that of your first circuit. Did increasing the length of wire in the circuit change the brightness of the bulb or the curr

Answers

Answer:

the resistance of the wire has no effect on the brightness of the bulb.

Explanation:

Let's apply ohm's law for your light bulb circuit plus wires plus switch

             V = I R_{bulb} + I R_ {wire}

the current in a series circuit is constant

             V = I (R_{bulb} + R_{wire})

To know the effect of the wires on the brightness of the bulb, we must look for the value of the typical resistance of these elements.

Incandescent bulb

Power 60 W

let's use the power ratio

            P = V I = V2 / R

            R = V2 / P

the voltage value for this power is V = 120 V

            R = 120 2/60

            R_bulb = 240 Ω

Resistance of a 14 gauge copper wire (most used), we look for it on the internet

            R = 8.45 Ω/ km

in a laboratory circuit approximately 2 m is used, so the resistance of our cable is

            R = 8.45 10⁻³ 2

            R_wire = 0.0169 Ω

let's buy the two resistors

            R_{bulb} = 240

            R_{wire} = 0.0169

            [tex]\frac{R_{bulb} }{R_{wire} } = \frac{240 }{ 0.0169}[/tex]

              [tex]\frac{ R_{bulb} }{ R_{wire} } = 1.4 \ 10^4[/tex]

therefore resistance of the bulb is much greater than that of the wire, therefore almost all the power is dissipated in the bulb.

In summary, the resistance of the wire has no effect on the brightness of the bulb.

A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 140 m above the launch point. (a) Calculate the horizontal component of its velocity. (b) Calculate the vertical component of its velocity just after launch. (c) At one instant during its flight the vertical component of its velocity is found to be 65 m/s. At that time, how far is it above or below the launch point

Answers

Answer:

a). 53.78 m/s

b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

In the c part, The negative distance is telling us that the project went below the lunch point.

2.Test the age of your eyes. a.Hold a pencil or ballpoint pen vertically at arm's length. b.Close your left eye and focus on the tip. c.Quickly bring the pencil closer to your eye until it is out of focus. d.Have your partner measure the distance between your eye and the pencil. e.Repeat for both eyes. f.Try it with and without glasses (if you wear glasses). Age of your Eyes Cm91013185083

Answers

Answer:

See Explanation

Explanation:

Given

Steps: a - f

Table

[tex]\begin{array}{ccccccc}{cm} & {9} & {10} & {13} & {18} & {50} & {83} \ \\ {Age} & {10} & {20} & {30} & {40} & {50} & {60} \ \end{array}[/tex]

Note that: The question is a practical question and the result may differ base on individuals and environment.

So, I will pick up the question from how to determine the age of the eye after the distance between the eyes and the pencil has been established

In my case, the measurement is:

[tex]Length= 10.4[/tex]

Approximate

[tex]Length= 10[/tex]

From the above table, the corresponding age to 10cm is:

[tex]Age = 20cm[/tex]

If in your measurement, the length is approximately (for example):

[tex]Length = 9cm[/tex]

The age will be:

[tex]Age = 10[/tex]

Suppose you are watching an ice-hockey game with some classmates. As the puck is sliding across the ice at a relatively constant speed, it passes close to a player, who hits it (with his stick) in the same direction as it is already moving. After the hit, you notice that the puck is moving much faster than it was before the hit. Your classmates propose the following three explanations for why the puck is moving faster after the hit than before. Whose reasoning do you agree with?
Student A: "The force of the stick was transferred to the puck during the hit. After the hit, the puck has more force, so it moves faster."
Student B: "The puck still has the force that started it moving, which is what keeps it moving at a constant speed. When the player hit it, he adds to this force, which makes the speed of the puck increase. After the hit, this extra force is gone, so the puck stops increasing its speed, but the original force is still there so now it moves at a faster constant speed."
Student C: "While the stick is in contact, it applies a force to the puck that makes its speed increase. As soon is contact is lost, this force disappears, so the speed of the puck stops increasing."

Answers

Answer:

the answer, the correct one is C

Explanation:

Let's propose the solution of this problem to know which explanation is correct, when the concha stick with the disc is an impulse exercise

                 

             I = ΔP

            ∫ F dt = pf-po

             ∫ F dt = m v_f - m v₀

Therefore, during the time that the contact lasts, a force is applied to the disk, which causes that if the amount of movement increases and therefore its speed increases, when the constant ceases the forces are reduced to zero and the disk no longer changes its momentum following with constant velocity.

When reviewing the answer, the correct one is C

A cat dozes on a stationary merry-go-round, at a radius of 7.0 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.9 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding

Answers

Answer:

0.6

Explanation:

Given that :

Radius, R = 7m

Period, T = 6.9s

The Coefficient of static friction, μs can be obtained using the relation :

μs = v² / 2gR

Recall, v = 2πR/T

μs becomes ;

μs = (2πR/T)² / 2gR

μs = (4π²R² / T²) ÷ 2gR

μs = (4π²R² / T²) * 1/ 2gR

μs = 4π²R / T²g

μs = 4π²*7 / 6.9^2 * 9.8

μs = 28π² / 466.578

μs = 276.34892 / 466.578

μs = 0.5922887

μs = 0.6

Match the following:​

Answers

iron metal :chromium

machinery part :nickel or chromium

ornamentation and decoration pieces :silver and gold

processed food :tin coated iron can

bridges and automobiles :zinc metal

distilled water:bad conductor

Answer:

iron metal :chromium

machinery part :nickel or chromium

ornamentation and decoration pieces :silver and gold

processed food :tin coated iron can

bridges and automobiles :zinc metal

distilled water:bad conductor

Explanation:

Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader

Answers

Answer:

7.55 g

Explanation:

Using the relation :

Δt = temperature change = (6° - 0°) = 6°

Q = quantity of heat

C = specific heat capacity = 4190 j/kg/k

1000 J = 1kJ

333 KJ = 333000 j

The quantity of ice that will melt ;

= 0.419 * 6 * 100 / 333000

= 2514000 / 333000

= 7.549 g

The mass of ice that will melt :

2.514 / 0.333

= 7.549 g

Car X is travelling at 30m/s north. Its driver looks at car Y approaching on another road and he estimates it is moving at 15m/s south-west relative to his car. Calculate the velocity of car Y relative to the ground.

Answers

Answer: 22.1 m/s

Explanation:

The velocity of Car traveling 30 m/s towards the north

In vector form it is

[tex]v_x=30\hat{j}[/tex]

The velocity of car Y w.r.t X is

[tex]\Rightarrow v_{yx}=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}][/tex]

Solving this

[tex]\Rightarrow v_{yx}=v_y-v_x\\\Rightarrow v_y=v_{yx}+v_x[/tex]

putting values

[tex]\Rightarrow v_y=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}]+30\hat{j}[/tex]

[tex]\Rightarrow v_y=-10.606\hat{i}+19.39\hat{j}[/tex]

absolute velocity relative to ground is

[tex]\left | v_y\right |=\sqrt{(-10.606)^2+(19.39)^2}\\\left | v_y\right |=22.101\ m/s[/tex]

Part D Here is one last question as a final check on your understanding of your work for this problem, looking at this problem as an example of the Conservation of Energy. The action in this problem begins at location A , with the block resting against the uncompressed spring. The action ends at location B, with the block moving up the ramp at a measured speed of 7.35 m/s . From A to B, what has been the work done by non-conservative forces, and what has been the change in the mechanical energy of the block-Earth system (the ramp is a part of the Earth)

Answers

Answer:

The answer is "39.95 J".

Explanation:

Please find the complete question in the attached file.

[tex]\to W_{AC}=(\mu \ m \ g \ \cos \theta ) d[/tex]

            [tex]=(0.45 \times 1.60 \times 9.8 \times \cos 26^{\circ}) 6.30 \\\\=(7.056 \times \cos 26^{\circ}) 6.30 \\\\=6.34189079\times 6.30\\\\=39.95 \ J\\\\[/tex]

[tex]\therefore \\\\\bold{\Delta E =39.95 \ J}[/tex]

Activities:
1. Name the instrument that is used to measure Air Pressure.
2.Explain what is Cyclone and Anticyclone

Answers

Answer: barometer.

A cyclone is a storm or system of winds that rotates around a center of low atmospheric pressure. An anticyclone is a system of winds that rotates around a center of high atmospheric pressure.

What is the formula for Ba+2 and F-?

Answers

Answer:

BaF2

Explanation:

since you got the valence numbers just do the scissors move where you:

give the F the 2 and give the Ba the 1 so it be like

BaF2 here is the chemical compound

Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 lb and was traveling eastward. Car B weighs 11251125 lb and was traveling westward at 42.042.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.517.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.7500.750 . How fast (in miles per hour) was car A traveling just before the collision

Answers

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]

we substitute the values

           v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

             

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

What is your hypothesis (or hypotheses) for this experiment?

(about Thermal Energy Transfer)

Answers

Answer:

I hypothesis that the motion involving the balls in the experiment were moving to create data.

Explanation:

I hope this helps!

A hot air balloon is rising at a speed of 10 km/hr. One hour later, the balloon
is still rising at 10 km/hr. What is its acceleration?

Answers

0
Acceleration is the rate of change of speed and there’s no change

Jim and Sally both do identical jobs. Jim works quickly while Sally works slowly. Which of the following is true?

A) Sally uses more energy.
B) Jim uses more energy.
C) Jim uses more power.
D) Sally uses more power.

Answers

The answer should be C

HELP PLEASE DUE IN 3 MINUTES

Answers

Answer:

Tectonic Plate Movement

Explanation:

Each continent and ocean sits on its own tectonic plate which floats on the Earths upper mantle. They move very little over time.

Answer:

tectonic plates movement

Which graph represents the relationship between the magnitude of the gravitational force exerted by earth on a spacecraft the distance between the center of the spacecraft the center of earth

Answers

Answer:

B as distance increase force decrease, but it is not a linear relationship.

A ball is dropped off the side of a bridge,
After 1.55 S, how far has it fallen?
(Unit=m)

Answers

Answer:

Distance S = 11.77 m (Approx.)

Explanation:

Given:

Time t = 1.55 Second

Gravity acceleration = 9.8 m/s²

Find:

Distance S

Computation:

S = ut + (1/2)(g)(t)²

S = (0)(1.55) + (1/2)(9.8)(1.55)²

S = (0)(1.55) + (1/2)(9.8)(1.55)²

Distance S = 11.77 m (Approx.)

Which one is it? Help ASAP

Answers

Answer:

extreme heat, because no physical damage can demagnetize a magnet

Explanation:

Answer:

the 3rd one

Explanation:

You have a hand-crank generator with a 100-turn coil (each turn having an area of 0.0350 m2), which can spin through a uniform magnetic field that has a magnitude of 0.0500 T. You can turn the crank at a maximum rate of 3 turns per second, but the hand crank is connected to the coil through a set of gears that makes the coil spin at a rate 6 times larger than the rate at which you turn the crank. What is the maximum emf you can expect to get out of this generator

Answers

Answer:

Explanation:

Maximum emf produced = nωAB. where n is no of turns , A is area , B is magnetic field and ω is angular velocity .

n = 100 , A = .035 m²

B = .05 T

ω = 2π f , f is no of revolution per second by coil

= 2 x 3.14 x 6 x 3

= 113.04 rad /s

Maximum emf produced = 100 x 113.04 x .035 x .05

= 19.78  volt .

plz help me with my career!!!
part one...

Answers

Answer:

#1 Yes

Explanation: #1: The rest of them are used mainley by farmers, and crops are used by common citizens in the world.

Question 1: Crops.

Question 2: Diagnostic Services.

Question 3: A cable company needs to lay new fiber optic cable to reach its customers across a large lake.

Question 4: A bachelor's degree in energy research.

Question 5: Environmental Resources.

If any of these answers are incorrect, please tell me, so I can fix my mistake. Thank you.

A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.

Answers

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2               ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1                  ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

In an effort to be the star of the half-time show, the majorette twirls a highly unusual baton made up of four mases fastened to the ends of light rods. Each rod is 1.0 m lone. Find the moment of inertia of the system about an axis perpendicular to the page and passing through the point where the rods cross.

Answers

Answer:

"0.25 kg-m²" is the appropriate answer.

Explanation:

The diagram of the question is missing. Find the attachment of the diagram below.

According to the diagram, the values are:

m₁ = 0.2

m₂ = 0.3

m₃ = 0.3

m₄ = 0.2

d₁ = d₂ = d₃ = d₄ = 0.5 m

As we know,

The moment of inertia is:

⇒  [tex]I=\Sigma M_id_i^2[/tex]

then,

⇒  [tex]I=m_1d_1^2+m_2d_2^2+m_3d_3^2+m_4d_4^2[/tex]

⇒     [tex]=d^2(m_1+m_2+m_3+m_4)[/tex]

On substituting the values, we get

⇒     [tex]=0.5^2\times (0.2+0.3+0.3+0.2)[/tex]

⇒     [tex]=0.25\times 1[/tex]

⇒     [tex]=0.25 \ Kg-m^2[/tex]

A spring in a toy gun has a spring constant of 10 N/m and can be compressed 4 cm.
It is then used to shoot a 1 g ball out of the gun. Find the velocity of the ball as it
leaves the gun

Answers

F=ma
We know from this that a=10,000m/s2

V=at
X=vt


You end up with v^2=ax

Plug in 10,000 and 0.04 and solve for v =20m/s
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