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2 years ago

Please write in complete sentences to answer the following:

Physics

1 answer:

ratelena [41]2 years ago

8 0

Answer:

Radio waves and the are various frequencies of radio waves are used for television and FM and AM radio broadcasts, military communications, mobile phones, ham radio, wireless computer networks, and numerous other communications applications. They are important because Radio waves are very widely used in modern technology for fixed and mobile radio communication, broadcasting, radar and radio navigation systems, communications satellites, wireless computer networks and many other applications.

Microwaves are widely used in modern technology, for example in point-to-point communication links, wireless networks, microwave radio relay networks, radar, satellite and spacecraft communication. They are important because microwaves play an increasingly wide role in heating and cooking food. They are absorbed by water and fat in foodstuffs (e.g., in the tissue of meats) and produce heat from the inside. In most cases, this reduces the cooking time a hundredfold.

Explanation:

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You look at a barometer and see that the air pressure is getting much higher... How will the weather probably change? *

Murljashka [212]

Answer:

The weather will clear up and get sunnier.

Explanation:As weather forecasters monitor air pressure, falling barometer measurements can signal that bad weather is on the way. In general, if a low pressure system is on its way, be prepared for warmer weather with storms and rain. If a high pressure system is coming, you can expect clear skies and cooler temperatures.

hope this helped:)

Brainliest?

8 0

3 years ago

Read 2 more answers

You have a string with a mass of 0.0127 kg. You stretch the string with a force of 9.33 N, giving it a length of 1.93 m. Then, y

melomori [17]

Answer:

wavelength = 0.968 m

frequency = 39.02 Hz

Explanation:

given data

mass = 0.0127 kg

force = 9.33 N

length = 1.93 m

to find out

wavelength and Frequency

solution

we know here linear density that is

linear density = $\frac{mass}{length}$ .........1

linear density = $\frac{0.0127}{1.93}$

linear density = 6.5803 × $10^{-3}$ kg/m

wavelength will be here

wavelength = $\frac{2L}{n}$ ..............2

here n = 4 for forth harmonic

wavelength = $\frac{2*1.93}{4}$

wavelength = 0.968 m

and

frequency will be for 4th normal mode of vibration is

frequency = $\frac{4}{2L} \sqrt{\frac{tension}{linear\ density} }$ ..........3

frequency = $\frac{4}{2*1.93} \sqrt{\frac{9.33}{6.5803*10^{-3}} }$

frequency = 1.036269 × 37.654594

frequency = 39.02 Hz

5 0

3 years ago

g typical coffee mug holds up to 12 oz (340 grams) of water. How much heat energy is required to warm enough water from 23⁰C to

Kryger [21]

Answer:

Explanation:

Heat required to raise the temperature

= mass x specific heat x rise in temperature

= .34 x 4200 x ( 95 - 23 )

= 102816 J .

1 kWh = 1000 x 60 x 60 J

= 3600000 J

102816 J = 102816 / 3600000

= .02856 kWh .

8 0

3 years ago

Meteor Infrasound A meteor that explodes in the atmosphere creates infrasound waves that can travel multiple times around the gl

ladessa [460]

Answer:

The frequency of these waves is $4.27\times10^{-2}\ Hz$

Explanation:

Given that,

Wavelength = 6.6 km

Distance = 8810 km

Time t = 8.67 hr

We need to calculate the velocity of sound

Using formula of velocity

$v = \dfrac{D}{T}$

Where, D = distance

T = time

Put the value into the formula

$v =\dfrac{8810}{8.67}$

$v=1016\ km/hr$

We need to calculate the frequency

Using formula of frequency

$v=n\lambda$

$n=\dfrac{v}{\lambda}$

Put the value into the formula

$n=\dfrac{1016}{6.6}$

$n=153.93\ hr$

$n=\dfrac{153.93}{60\times60}$

$n=0.0427\ Hz$

$n=4.27\times10^{-2}\ Hz$

Hence, The frequency of these waves is $4.27\times10^{-2}\ Hz$

8 0

3 years ago

If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?

Arisa [49]

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm$

Magnification,

$m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25$

Hence, the magnification of the image is 0.25.

7 0

3 years ago