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Maksim231197 [3]
2 years ago
9

1. Discuss how we use trial and error, algorithms, heuristics, and insight to solve problems. For each concept, define the term

and then describe an incident (either from your own experience or that of a friend) in which the concept was used to determine a solution to a specific problem. You must use specific details of the incident and terms from our module to show me you understand the concept. (Total 64 points)
2., How can confirmation bias and fixation can interfere with effective problem solving. In this portion of your response, for each concept, define the term and then describe an incident (either from your own experience or that of a friend) in which the concept interfered with an appropriate solution being determined. You must use specific details of the incident and terms from our module to show me you understand the concept. (Total 32 pts).
Physics
1 answer:
IgorC [24]2 years ago
7 0

Answer: used concepts to simplify and order the world around us, categories: objects, events, ideas, or people

hierarchies: subdivide categories into smaller more detailed units, prototypes: best example

algorithm: time-consuming but thorough set of rules or procedures

heuristics: simpler way of thinking, solve problems but maybe incorrect solutions

insight: flash of inspiration that solves problem

representativeness heuristic: judge likelihood of things in terms of how they represent our prototype

availability heuristic: judge likelihood of things based on how vivid they are or how readily they come to mind.

Explanation: i hope that helped!

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Question 15 (3.33 points) Solve: What work is done when 3.0 C is moved through an electric potential difference of 1.5 V?
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A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
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Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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2 years ago
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