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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.

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A student is asked to describe the path of a paper airplane that is thrown in the classroom. Which statement best describes the
emmasim [6.3K]

Answer: The paper airplane will create a curved path towards the floor as it is pulled toward <u><em>Earth's center.</em></u>

Explanation: The paper airplane will be pulled to the center because <u><em>Earth has a much greater mass than objects on its surface.</em></u> And it will curve because of the amount of <u><em>force</em></u> you are putting onto the plane.

4 0
2 years ago
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If a pizza delivery guy declares himself the leader over the other pizza delivery people, he probably won't have many people obe
NemiM [27]

Answer:

the last one i think...

4 0
1 year ago
13. An aircraft heads North at 320 km/h rel:
AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

v' = v + v_a

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

v_a is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

v_a = -80 km/h (since the air is moving from North)

Therefore, we find

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

brainly.com/question/4945130

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7 0
2 years ago
A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity
Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

\omega_i = 2\pi f_1

\omega_1 = 2\pi(\frac{610}{60})

\omega_1 = 63.88 rad/s

similarly final angular speed is given as

\omega_f = 2\pi f_2

\omega_2 = 2\pi(\frac{837}{60})

\omega_2 = 87.65 rad/s

angular acceleration of the turbine is given as

\alpha = 5.9 rad/s^2

now we have to find the angular displacement is given as

\theta = \omega t + \frac{1}{2}\alpha t^2

\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)

\theta = 234.62 radian

3 0
2 years ago
A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to
Annette [7]

Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.

In order to do so we can use the following equation:

y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

0=y_{0}+\frac{1}{2}at^{2}

and solve for t:

t=\sqrt{\frac{-2y_0}{a}}

now we can substitute the values:

t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the  distance from the ship to point A to be:

x_{A}=V_{x}t

x_{A}=(100m/s)(3.19)

x_{A}=319m

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}

So we can substitute the given values:

x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}

Which yields:

x_{f}=329.18m

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

So the shell missed the buoy by 10.18m.

8 0
2 years ago
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