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3 years ago

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Which model of communication was originally described as a one-way process consisting of the following five questions: who, says

what, in what channel, to whom, and with what effect?

1 answer:

tresset_1 [31]3 years ago

5 0

Answer:

Lasswell's 5 step model

Explanation:

This model of communication defines the answers to the questions who says, to whom, says what, the effects it was said with and in what channel.

Lasswell's 5 step model of communication was described as a single way process and was used for the analysis of mass communication, communication in groups and between people to disperse messages into these. It was designed for the purpose of studying and analyzing the business strategies and the media hype of different countries at that time.

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In two or more complete sentences, explain how you can prove that the number of degrees that the Moon rotates around the Earth e

sasho [114]

A close representation of the moons motion around earth is that it moves around the earth once in 27.3 days. the moons average movement is 13.2 degrees per day so around 92 per week.

Hope this helps!

5 0

4 years ago

A bystander observes the musicians heading toward each other. When musician #1 is 100 m away, the intensity is 1.24 x 10-8 W/m^2

777dan777 [17]

Explanation:

Given that,

Distance 1, r = 100 m

Intensity, $I_1=1.24\times 10^{-8}\ W/m^2$

If distance 2, r' = 25 m

We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :

$I=\dfrac{P}{4\pi r^2}$ .........(1)

Let I' is the intensity at r'. So,

$I'=\dfrac{P}{4\pi r'^2}$ ............(2)

From equation (1) and (2) :

$I'=\dfrac{Ir}{r'^2}$

$I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}$

$I'=1.98\times 10^{-9}\ W/m^2$

Intensity level is given by :

$dB=10\ log(\dfrac{I'}{I_o})$ , $I_o=10^{-12}\ W/m^2$

$dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})$

dB = 32.96 dB

Hence, this is the required solution.

7 0

3 years ago

Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0

4 years ago

9- Under what circumstances would a vector have components that are equal in

valkas [14]

Explanation:

c. if the vector is oriented at 0° from the X -axis.

6 0

2 years ago

which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given

Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>

Weight of Punch on Earth = 236 lb
Desired weight = 118 lb

The mass of Punch will be constant in every planet;

$W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}$

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

$F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}$

where;

M is the mass of Earth = 5.972 x 10²⁴ kg
R is the Radius of Earth = 6,371 km

For Planet Tehar;

$g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g$

For planet Loput:

$g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g$

For planet Cremury:

$g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g$

For Planet Suven:

$g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g$

For Planet Pentune;

$g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g$

For Planet Rams;

$g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g$

The weight Punch on Each Planet at a constant mass is calculated as follows;

$W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb$

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

<u>The </u><u>complete question</u><u> is below</u>:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).

<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>

Learn more about effect of gravity on weight here: brainly.com/question/3908593

5 0

2 years ago