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scZoUnD [109]
4 years ago
9

A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i

s the best approximation of the box’s acceleration?
Physics
2 answers:
antiseptic1488 [7]4 years ago
8 0
Fnet=m*a
a= Fnet/m

a=(20N[E]+9.0N[E])/ 12.0kg

a=2.4167m/s^2

a ≈ 2.4m/s^2
wariber [46]4 years ago
5 0

Answer:

The acceleration of the box its 0,91 m/s²

Explanation:

According to <em>Newton</em> first law, the force applyed to some object its equal to the mass of the object plus the acceleration of the same object.

   F=m.a ; clearing the acceleration of the equation ⇒ a=F/m

  Now in the problem its apeared the friction force and according to de phsysics definition its a force thats its against of movement.

  So the force its going to be equal to F= 20N - 9N.

 <u>Replacing in the equation</u> a=(20N - 9N)/12kg.

 <u>Solving</u> ⇒ a ≅0, 91 m/s²

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3 years ago
As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length
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Answer:

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is Lw. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

Question:

If the bottom of your window is a height hb above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb, the speed at the bottom of the window, defined by

vb=Lwt+gt2.

Express your answer in terms of some or all of the variables hb, Lw, t, vb, and g.

The correct answer is

vground = (t²+lw²+(2t³+2t²)×g×Lw+(t⁴+2t³+t²) ) ∧ 0.5 = vb² +2g×vbt+1/2gt²

Explanation:

We have from the relation

v = u + gt, S = ut + 1/2 gt², v² = u² + 2gS

in this case S = hb, u = vb=Lwt+gt2.  and v = vground

therefore v² = (Lwt+gt²)² + 2 × g × hb

= (Lwt+gt²)² + 2 × g ×  (Lwt+gt²)×t + 1/2 gt² = vb² +2g×vbt+1/2gt²

vground = (t²+lw²+(2t³+2t²)×g×Lw+(t⁴+2t³+t²) ) ∧ 0.5

7 0
3 years ago
A hiker walks due east for a distance of 25.5 km from her base camp. On the second day, she walks 41.0 km northwest till she dis
GarryVolchara [31]

Resultant displacement is 29.2 km at 83.1^{\circ} north of west

Explanation:

To solve the problem, we have to use the rules of vector addition, resolving first each vector along the x- and y- direction.

Taking east as positive x direction and north as positive y- direction, we have:

- First displacement is 25.5 km east, therefore its components are

A_x = 25.5 km\\A_y = 0 km

- Second displacement is 41.0 km northwest, so its components are

B_x = (41.0)cos(135^{\circ})=-29.0 km\\B_y =(41.0)sin(135^{\circ})=29.0 km

So, the components of the resultant displacement are

R_x=A_x+B_x=25.5+(-29.0)=-3.5 km\\R_y=A_y+B_y=0+29.0=29.0 km

And so, the magnitude is calculated using Pythagorean's theorem:

R=\sqrt{R_x^2+R_y^2}=\sqrt{(-3.5)^2+(29.0)^2}=29.2 km

And the direction is given by

\theta=tan^{-1}(\frac{R_y}{|R_x|})=tan^{-1}(\frac{29.0}{3.5})=83.1^{\circ}

Where the angle is measured from the west direction, since Rx is negative.

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

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