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Ymorist [56]
3 years ago
12

Bianca sends her brother Phineas an e-mail to invite him to dinner. In this scenario, what is the channel according to Shannon's

Model of Communication?
A. Bianca
B. dinner
C. e-mail
D. Phineas
need help
Physics
1 answer:
dusya [7]3 years ago
4 0
The channel is whatever was used to carry information from
one place to another, or from one person to another.
In this scenario, the channel is e-mail.
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A cabbie is trying to stop when he notices a fare is whistling them over. The
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  • K.E=18750J
  • Mass=m=2100kg
  • Velocity=v

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2

\\ \sf\longmapsto 18750=1050v^2

\\ \sf\longmapsto v^2=\dfrac{18750}{1050}

\\ \sf\longmapsto v^2=17.85m^2

\\ \sf\longmapsto v=\sqrt{17.85}

\\ \sf\longmapsto v=4.1m/s

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In an experiment, a researcher can make claims about causation
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<span>In an experiment, a researcher can make claims about causation if the independent variable changes because of changes made to the dependent variable. Causation works on cause and effect, so the changed independent variable is the cause and the changed dependent variable is the effect. In an experiment the independent variable is changed to determine the dependent variables value, so the two are directly related.</span>
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3. Jeremy has a solution with a mass of 228 g. He knows that the mass of the solvent was 224 g. How much is the mass of the solu
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The period of a sound wave is 0.002 seconds. The speed of a sound is 344 m/s. Find the frequency and wavelength of the sound wav
ankoles [38]

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5 0
3 years ago
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

5 0
3 years ago
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