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alekssr [168]
2 years ago
13

Torque, a car manufacturer, aims to make its slogan "Burn Rubber" known to every individual in a way that they immediately assoc

iate it with the firm. The company has created a social media page where almost a thousand fans have registered. However, there are two competitors who advertise the same features as those of Torque cars. The firm has a large number of customers who think it manufactures the best cars and there is no other manufacturer better than Torque.
Physics
1 answer:
mylen [45]2 years ago
7 0

Answer:

A customer is about to buy a limited edition sports car from Torque. It is most likely that the customer will have  d. limited problem solving, in which consumers decision rules to purchase are simple, and are not motivated to search for information about other optons, deciding to buy the car.

Explanation:

a. advertising clutter is the big amount of ad-messages that the consumer is exposed to everyday.

b. high involvement  product or purchases features many variables to be considered by the customer before getting to a decision.

c. cognitive dissonance  or conflicting attitudes causing mental discomfort.

You might be interested in
If the volume of an object is reported as 5.0 ft3 what is the volume in cubic meters
12345 [234]
The problem statement is simply asking us to convert units. We convert from units of ft^3 to units of m^3. To do this, we need a conversion factor. For this case, we use 1 m is equal to 3.28084 ft. We do as follows:

5.0 ft^3 ( 1 m / 3.28084 ft )^3 = 0.1416 m^3
3 0
2 years ago
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0
2 years ago
A spring-powered dart gun is unstretched and has a spring constant 16.0 N/m. The spring is compressed by 8.0 cm and a 5.0 gram p
stepladder [879]

Answer:

Explanation:

Given that,

Spring constant = 16N/m

Extension of spring

x = 8cm = 0.08m

Mass

m = 5g =5/1000 = 0.005 kg

The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.

So, Using conservation of energy

Energy in spring is converted to kinectic energy

So, Ux = K.E

Ux = ½ kx²

Then,

Ux = ½ × 16 × 0.08m²

Ux = 0.64 J

Since, K.E = Ux

K.E = 0.64 J

4 0
1 year ago
At what distance above earth would a satellite have a period of 125 min?
Nezavi [6.7K]
Rw^2 = GmM/r^2 
<span> Leads to 
</span><span> w^2 r^3 = GM 
</span><span> (2pi /T) ^2 r^3 = GM 
</span><span> 4pi^2 r^3 = GM T^2 
</span><span> r^3 = GM T^2 / 4pi^2 
</span><span> Work out r^3 then r. 
</span> T = 125 min = 125(60) = 7500 s 
<span> R = 6.38E6 m 
</span><span> m = 5.97E24 kg 
</span><span> G = 6.673E-11 
</span> r=<span> 8279791.78</span><span> m
 Since r = radius R of Earth + height above urface,h 
</span><span> h = r - R = </span><span> 8279791.78 - </span>6.38E6 = <span> <span>1899791.78 m
 h=</span></span><span> <span>1899.79178 Km</span></span>
5 0
1 year ago
Read 2 more answers
The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f
zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

                 584 = Ha + Lg

an also this mean that torques acting on the body is balanced

         So,   0.410 Ha  = 0.840 Lg

    Making Lg the subject of formula in the equation above we

   Lg = 0.4881 Ha

 Considering the first equation and replacing Lg with this recent equation we have

                      584 = Ha + 0.4881 Ha

          Therefore Ha = 392.44 N

This value obtained is  for both hands for each hand we divide by 2

Therefore we have for each hand = 392.44/2 =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

             Lg = 0.4881 ×  392.44

                  = 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

                  191.22/2 = 95.8 N.

8 0
2 years ago
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