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krok68 [10]
2 years ago
14

Click to review the online content. Then answer the question(s) below, using complete sentences. Scroll down to view additional

questions.
Site 1: "Facial Feedback Theory Script"

After watching the case study and reading the script about the facial feedback theory, reflect on how you feel. Did you realize that none of the images showed angry faces? Why do you think you viewed only pictures of smiling people, and what effect do you think it was meant to have on you as the viewer? What do you think would occur if you were asked to view the case study again, this time using a mirror to mimic the faces you saw on screen? (Site 1)
Physics
1 answer:
vivado [14]2 years ago
6 0

The facial feedback theory is a psychological theory associated with the emotional consequences due to physiological actions.

The provided case study a number of smiling faces which made the person saying it smile thereby working there muscles that related with smile this makes a psychological consequence and people find  things much  more funnier than they usually are.

If as a viewer again this case study is provided and the viewers is allowed to mimic in mirror he or she will again make much  funnier faces due to the psychological consequence.

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The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2
sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed V_{0}

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}

Once we have the initial speed, we can isolate the final speed from following equation:

V_{f} =V_{0} +a*t  V_{f}= 4.165 \frac{m}{s}  

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }

3 0
2 years ago
Read 2 more answers
Your town is installing a fountain in the main square. If the water is to rise 26.0 m (85.3 feet) above the fountain, how much p
Brums [2.3K]

Answer:

P = 3.55 \times 10^5 Pa

Explanation:

As we know that water from the fountain will raise to maximum height

H = 26.0 m

now by energy conservation we can say that initial speed of the water just after it moves out will be

\frac{1}{2}mv^2 = mgH

v = \sqrt{2gH}

v = \sqrt{2(9.81)(26)}

v = 22.6 m/s

Now we can use Bernuolli's theorem to find the initial pressure inside the pipe

P = P_0 + \frac{1}{2}\rho v^2

P = 10^5 + \frac{1}{2}(1000)(22.6^2)

P = 3.55 \times 10^5 Pa

6 0
2 years ago
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m  

0.80 m  

1.25 m

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

8 0
2 years ago
Read 2 more answers
A man swims at a speed of 0.4 m/s. How long will it take him to cross a pool of length 50 m?
konstantin123 [22]
Distance = speed * time, then time = distance / speed.

time = 50 / 0.4 = 125 s
7 0
2 years ago
List some reasons why growth characteristics are more useful on agar plates than on agar slants
SpyIntel [72]
Usually, in culturing of the bacteria we have a slant and then portion f it is transferred to the agar plate. The growth characteristics are more useful in the agar plates because it is where we really do the observation because bacteria in slants are still to be transferred in the agar plates. 
5 0
2 years ago
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