We can find the y-component of the resultant force by adding the y-components of the two 20N forces.
For a force of magnitude F and lying at an angle off the x-axis θ, the y-component of the force is given by:
Fsin(θ)
The magnitude of the two forces is 20N, and they lie at 30° and 60°, so the sum of their y-components, and therefore the y-component of the resultant force, is:
20sin(30°)+20sin(60°)
= 27.3N
Phase 1. Forethought/preaction—This phase precedes the actual performance; sets the stage for action; maps out the tasks to minimize the unknown; and helps to develop a positive mindset. Realistic expectations can make the task more appealing. Goals must be set as specific outcomes, arranged in order from short-term to long-term. We have to ask students to consider the following:
<span>When will they start?Where will they do the work?How will they get started?<span>What conditions will help or hinder their learning activities are a part of this phase?
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Phase 2. Performance control—This phase involves processes during learning and the active attempt to utilize specific strategies to help a student become more successful.
We have to ask students to consider the following:
<span>Are students accomplishing what they hoped to do?Are they being distracted?Is this taking more time than they thought?Under what conditions do they accomplish the most?What questions can they ask themselves while they are working?<span>How can they encourage themselves to keep working (including self-talk—come on, get your work done so you can watch that television show or read your magazine!)
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Phase 3. Self-reflection—This phase involves reflection after the performance, a self-evaluation of outcomes compared to goals.
We have to ask students to consider the following:
<span>Did they accomplish what they planned to do?Were they distracted and how did they get back to work?Did they plan enough time or did they need more time than they thought?<span>Under what conditions did they accomplish the most work.
Hope this helps!!!!!
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Answer:
a = 0.8 [m/s²]
Explanation:
To solve this problem we must use Newton's second law which tells us that the resulting force on a body is equal to the product of mass by acceleration, in this way we come to the following equation:
∑F = m*a
where:
F = forces applied [N]
m = mass = 1000 [kg]
a = acceleration [m/s²]
Now using Newton's second law.
![1200 - 400 = 1000*a\\800 = 1000 *a\\a=0.8[m/s^{2} ]](https://tex.z-dn.net/?f=1200%20-%20400%20%3D%201000%2Aa%5C%5C800%20%3D%201000%20%2Aa%5C%5Ca%3D0.8%5Bm%2Fs%5E%7B2%7D%20%5D)
Answer:
The mass of the second person is 28.91 kg
Explanation:
Given;
mass of the first adult, m₁ = 54.2 kg
distance of the first adult from the point of balance, x = 1.20 m
mass of the second adult, = m₂
distance of the second adult from the point of balance, y = 2.25 m
Taking moment about the point of balance, we will have
m₁x = m₂y
54.2 x 1.2 = 2.25y
2.25y = 65.04
y = 65.04/2.25
y = 28.91 kg
Therefore, the mass of the second person is 28.91 kg
The rate at which someone or something can move or operate very fast.
