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4 years ago

15

Look at the advertisement. A poster that reads, "There's more to resistors than resistance. If you're really serious about cost,

be serious about quality." with tools. Which persuasive media technique is used in this advertisement

2 answers:

ozzi4 years ago

7 0

Options:

universal appeal

flattery

association

bandwagon

Answer: Association

Explanation: Advertisement is a marketing technique through which business organisations utilize the opportunities provided by both the print, electronic and other channels of communication to market a product to the target audience.

Association Advertising is a type of Advertising where certain attributes which have been known to be associated with good and quality products are used to market the products to the target audience.

nadezda [96]4 years ago

7 0

Explanation:

It's Flattery guys

You might be interested in

How can you determine the rate of a chemical reaction?

DanielleElmas [232]

Take a look at a simple reaction like the one below:

In this reaction some reactant A is turned into some product B. The rate of reaction can be represented by a decrease in concentration of A over time or as the increase of B over time. This is written:

6 0

3 years ago

A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons

shepuryov [24]

Answer:

$v=2.556m/s$

Explanation:

From the conservation of mechanical energy

$K_{E1}+U_1=K_{E2}+U_2$

$\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2$

$x_2=0.08m$

$v_1=0 m/s$

Solve to velocity v2

$m*v_2^2=k*x_1^2-k*x_2^2$

$v^2=\frac{k}{m}*(x_1^2-x_2^2)$

$v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)$

$v=\sqrt{6.54m^2/s^2}=2.556m/s$

4 0

3 years ago

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it i

lianna [129]

Answer:

<em>A) the moment of inertia of the system decreases and the angular speed increases. </em>

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

$I_{1} w_{1} = I_{2} w_{2}$ ....1

where $I_{1}$ and $I_{2}$ are the initial and final moment of inertia respectively.

and $w_{1}$ and $w_{2}$ are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

$I = mr^{2}$ ....2

where $m$ is the mass of the rotating body,

and $r$ is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from $I_{1}$ to $I_{2}$ will cause the angular speed of the system to increase from $w_{1}$ to $w_{2}$ .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.

7 0

4 years ago

A cube of wood having an edge dimension of 20.0cm and a density of 650 kg /m³ floats on water. (a) What is the distance from the

Zarrin [17]

The distance from the horizontal top surface of the cube to the water level is "6.282 cm".

<h3>What is Archimedes' principle?</h3>

According to Archimedes' principle, the weight of the fluid that the body displaces is equal to the upward buoyant force that is applied to a body submerged in a fluid, whether fully or partially. The Archimedes' principle is a fundamental physical law in fluid mechanics. It was created by Syracuse's Archimedes.

According to Archimedes' principle, a body submerged in a fluid experiences an upward force proportional to the weight of the fluid that has been displaced. One of the prerequisites for equilibrium is this. We believe that the buoyancy force, also known as the centre of buoyancy, is situated in the middle of the submerged hull.

From Archimedes' principle, we get

$\rightarrow L^3 \rho_{\text {Wood }} &=L^2 d \rho_{\text {Water }} \\$

$d &=L \frac{\rho_{\text {Waat }}}{\rho_{\text {Water }}} \\$

$&=18 \times \frac{651}{1000} \\$

=11.72cm

So,

The distance from horizontal top to the water level will be:

=18-11.72

=6.282cm

To learn more about Archimedes' principle refer to:

brainly.com/question/1155674

#SPJ4

4 0

1 year ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f

Alborosie

Answer:

3.25 m

Explanation:

t = Time taken = 0.166 seconds

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s²

s = 1 because meter stick is 1 meter in length

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s$

Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m$

The distance of the acorn from the ground is 1.87+1.38 = 3.25 m

4 0

3 years ago