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xxTIMURxx [149]
2 years ago
8

Two bulldozers are pushing on a large tree trunk as shown. (last photo)

Physics
1 answer:
andrey2020 [161]2 years ago
8 0

The tree will fall in the south eastern side of the area due to two bulldozer.

<h3>Direction of tree falling</h3>

The tree falls in the sloppy direction due to both bulldozer which apply force from two directions. One bulldozer apply force on the tree from western side whereas the other bulldozer apply force on the tree from north side so that's why we can say that the tree will fall in the south eastern side of the area.

Learn more about force here: brainly.com/question/12970081

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What is potential energy
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Answer:

Potential energy is energy that is stored – or conserved - in an object or substance. This stored energy is based on the position, arrangement or state of the object or substance. You can think of it as energy that has the 'potential' to do work.

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Explain how forces of attraction and repulsion exist within an atom
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Answer:

Oppositely charged particles attract each other, while like particles repel one another. Electrons are kept in the orbit around the nucleus by the electromagnetic force, because the nucleus in the center of the atom is positively charged and attracts the negatively charged electrons.

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An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
Aneli [31]

Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

We have

\displaystyle v_o=20\ m/s,\ \theta=25^o

The maximum height is

\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}

\displaystyle y_m=3.65m

7 0
2 years ago
The path followed by a projectile is called its _____.
xxTIMURxx [149]

The path followed by a projectile is called its <em>trajectory. (C)</em>

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castortr0y [4]

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