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xxTIMURxx [149]
3 years ago
6

An apple is pushed across a table with a velocity of 3.74 m/s rolls too far and falls 0.89

Physics
1 answer:
Tcecarenko [31]3 years ago
8 0

The apple lands 1.61 m from the base of the table

Explanation:

The motion of the apple here is a projectile motion, therefore it consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

First we analyze the vertical motion, to find the time of flight of the apple. We can do it by using the following suvat equation:

s=u_y t+\frac{1}{2}at^2

where, choosing downward as positive direction,  we have:

s = 0.89 m is the vertical displacement of the apple

u_y=0 is the initial vertical velocity  (since it is thrown horizontally)

t is the time of the fall

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t,  we find

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(0.89)}{9.8}}=0.43 s

So the apple takes 0.43 s to fall to the ground.

Now we can analyze the horizontal motion: the apple moves horizontally with a constant velocity of

v_x = 3.74 m/s  (its initial velocity)

Therefore, the distance it covers in a time t is given by

d=v_x t

and by substituting t = 0.43 s, we find the final distance of the apple from the base of the table:

d=(3.74)(0.43)=1.61 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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In a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0° above the horizon
Masja [62]

Answer:38.66 m

Explanation:

Given

launch angle \theta =51^{\circ}

launch velocity u=18 m/s

center of mass continue to travel its original Path so it center of mass will be at a distance of

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You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

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It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

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Learn more about the direction of the force here,

brainly.com/question/2037071

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