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2 years ago

6

Two ways that radio waves are used for transmitting information. Help me please.

1 answer:

Travka [436]2 years ago

3 0

1. Stop transmitting start and stop sending photons. This is typical morse code like transmissions of information. By varying the pattern of starts stops you vary the information sent. Also used in digital communications over radios, modems, and pulse dialing over radio/phones.

<span>2. Change the frequency. You use a pattern of shorter and longer wavelengths of radio waves higher and lower energies of individual photons to transmit information. This is frequency modulation </span>

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You are holding a shopping basket at the grocery store with two 0.55-kg cartons of cereal at the left end of the basket. the bas

stepan [7]

Refer to the diagram shown below.

The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.

The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N

The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N

The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.

For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m

Answer: 0.63 m from the left end.

5 0

2 years ago

A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the

Agata [3.3K]

Answer with Explanation:

We are given that

Mass of squirrel,m=530 g= $\frac{530}{1000}=0.530 kg$

1kg=1000 g

Area=A= $935 cm^2=935\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Height,h=4.4 m

C=1

$\rho=1.21 kg/m^3$

Width of rectangular prism,b=11.6 cm= $\frac{11.6}{100}=0.116 m$

1 m=100 cm

Length,l=23.2 cm= $0.232 m$

Area= $l\times b=0.116\times 0.232=0.0269 m^2$

Terminal velocity, $v_t=\sqrt{\frac{2mg}{\rho CA}}$

Where $g=9.8 m/s^2$

Using the formula

$v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}$

$v_t=17.86 m/s$

The velocity of person, $v=\sqrt{2gh}$

Using the formula

$v=\sqrt{2\times 9.8\times 4.4}$

$v=9.29 m/s$

4 0

2 years ago

PLZZ HELP WILL MARK BRAINLY The function of the respiratory system is to ___________.

pickupchik [31]

Breathe and now I’m just filling in more letters so it’ll go thru

3 0

3 years ago

A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv

sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

$K_{rel}=(\gamma-1)mc^2$

$\dfrac{K_{rel}}{mc^2}=\gamma-1$

Put the value into the formula

$\gamma=\dfrac{105.7}{0.511}+1$

$\gamma=208$

We need to calculate the electron’s velocity

Using formula of velocity

$\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}$

$\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}$

$\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1$

$v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2$

Put the value into the formula

$v^2=\dfrac{1-(208)^2}{-208^2}\times c^2$

$v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}$

$v=0.9999 c\ m/s$

Hence, The electron’s velocity is 0.9999 c m/s.

6 0

2 years ago

NASA launches a probe with a mass of 15,000 kg to another planet more massive than Earth. Which statement is true about the prob

vivado [14]

Answer: The weight of the probe will increase.

8 0

2 years ago