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alex41 [277]
3 years ago
15

Peer assessment is a unique educational model. Think back to how you felt about peer assessment at the beginning of the term, an

d compare that to your feeling now. How have your feeling changed? Are you more comfortable with peer assessment? Have you learned something new while assessing your peer's work?​
Physics
2 answers:
luda_lava [24]3 years ago
8 0

Answer:

In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.

stepan [7]3 years ago
5 0

Answer:

He is right!

Explanation:

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Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y
Gnesinka [82]
The representation of this problem is shown in Figure 1. So our goal is to find the vector \overrightarrow{R}. From the figure we know that:

\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}

From geometry, we know that:

\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}

Then using vector decomposition into components:

For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85

Therefore:

R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
</span>
\left | \overrightarrow{R} \right |=&#10;\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


6 0
3 years ago
A group of students is measuring the change in momentum of an object. Which set of lab equipment would be most in measuring the
a_sh-v [17]

Answer:

The momentum of an object is defined as the mass of the object times the velocity of the object, as P = m*v.

So the equipment needed would be:

Something to measure the mass of the object, like a balance.

Something to measure the speed of the object, like a doppler radar, or a simpler thing may be a cronometer, with that you can measure the amount of time that the object needs to travel a given distance, and with that you can obtain the speed of the object.

Now you can notice that speed is different than velocity, this is true, velocity is a vector, so this has a direction, then you need something to fix the direction in which the object moves, in this way you can determine the velocity.

4 0
3 years ago
6/10
g100num [7]

Answer:pounds

Explanation:

7 0
3 years ago
A machine part has the shape of a solid uniform sphere of mass 205 g and diameter 4.10 cm. It is spinning about a frictionless a
kondor19780726 [428]

Answer:

Answer is in the following attachment.

                                                                                                                                                                                                 

Explanation:

3 0
3 years ago
If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?
andreyandreev [35.5K]

Answer:

v = -6m/s

Explanation:

x=4-12t+3t^2

\frac{dx}{dt}=-12+6t

For t = 1:

\frac{dx}{dt}=-6

3 0
3 years ago
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