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2 years ago

PLEASE HELP :)

1 answer:

6 0

1) in the opposite direction always
2) drag force is the force that acts in the opposite direction of the applied force (air resistance if a drag force for example)
3) frictions forces between the sole and the ground as the owner walks
4) chalk would glide very easily, so a creaking sound could never be produced, but also one could not write anything on the board!
5) I apologize for my lack of familiarity with the kind of sport but the probable answer is to raise the friction of their hands in order to decrease slipping and get a tighter, more firm grip

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because it can be hard

Explanation:

I said that because they be on bed rest

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3 years ago

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 how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:

$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the 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:

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

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3 years ago

Starting at the same time, an arrow and a ball are shot horizontally with a speeds of 50 m/s and 44 m/s respectively from the to

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Answer:

option D

Explanation:

given.

horizontal velocity of arrow and a ball given as 50 m/s and 44 m/s respectively from the top of a building over flat ground.

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they will react bottom at the same time

time of flight is same for both

now,

In horizontal direction,

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the correct answer is option D

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The sum of potential energy and kinetic energy.
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Explanation:

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2 years ago

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