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2 years ago

Remote controls rely on ____waves to carry signals and transmit information over short distances.

2 answers:

7 0

The small infrared LED on the top of a typical remote control. ... Most remote controls send signals using infrared radiation (which is a kind of invisible red light that hot objects give off and halogen hobs use to cook with), though some use radio waves instead.

i would go with infrared radiation.

please vote brainliest if this helped! :)

Likurg_2 [28]2 years ago

5 0

Answer:

Infrared Waves

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A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

Sedbober [7]

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ -- (A)

here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg 

Plug in the values in Equation (A)

so $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ = 197.97Hz 

the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
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2 years ago

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Metals are used in many products because of the characteristic properties that most metals have. Which product requires the high

labwork [276]

The answer is mirrors. Mirrors are made by applying a metal thin layer on the back surface of a transparent substrat, typically glass. The metal layer in the antiquity was bronze, mercury and later silver whose luster gave the reflective property to the mirror.

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2 years ago

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The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:

ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

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2 years ago

A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto

daser333 [38]

Convert the shaft ouput from HP to kW

Shaft output = 75HP = 55.93kW

1st: Finding for the power consumption based on 55.93kW output

Power consumption (Old) = 55.93kW / .910 = 61.46kW

Power consumption (New) = 55.93kW / .954 = 58.63kW

2nd: Total power used in kWh:

Power Used = Power consumption * load factor * Hours:

Power (Old) = 61.46kW * .75 * 4368 = 201343 kWh

Power (New) = 58.63kW * .75 * 4368 = 192072 kWh

Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

3rd: Calculating for the price:

Price = kW-Hr * $/kWh

Price (Old) = 201343kWh * $0.08/kWh = $16107.44

Price (New) = 192072 kWh * $0.08/kWh = $15365.76

Cost saved = $16107.44 - $15365.76 = $741.68/yr

4th: Setting up the cost equation:

Cost over time, F(t) = Motor_Cost + (Price * Number of Years, t)

Cost (Old) = 5449 + 16107.44*t

Cost (New) = 5520 + 15365.76*t

Equate the two to find for t when they cost equally:

5449 + 16107.44*t = 5520 + 15365.76*t

16107.44*t = 15365.76*t +71

16107.44*t - 15365.76*t = 71

741.68*t = 71

t = 71 / 741.68 = .095 years = 35 days

So the payback period is after 35 days.

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2 years ago

The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If th

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<h2>Answer:</h2>

The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:

1. (There would be variables that would be hard to control, leading to less reliable data.)

3. (Meter sticks may lack precision or may be read incorrectly.)

4. (Real glider data may vary since real collisions may involve loss of energy.)

5. (Human error in recording or plotting the data could be a factor.)

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2 years ago

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