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Alex777 [14]
3 years ago
13

When you flick a coin so that it slides across a table, what four forces are acting on the coin?

Physics
2 answers:
Kamila [148]3 years ago
8 0

<u>Any time the coin is on the table:</u>
-- gravity acting downwards
-- normal force of the table, acting upwards

<u>Any time the coin is moving:</u>
-- friction force, on the coin's underside
-- air resistance
Both of these act opposite to the direction of the coin's motion.

<u>During the flick, while your finger is still touching the coin:</u>
-- the flicking force, a push exerted by your finger
   The force ends as soon as the coin leaves contact with your finger.

Nadusha1986 [10]3 years ago
3 0
Pushing force
friction
gravity
normal force
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From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the
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The launching point is at a distance D = 962.2m and H = 39.2m

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It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

           x = Vox t

           t = x / vox

           t = 7.1 / 340

           t = 2.09 10-2 s

In this same time the height of the window fell

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Let's calculate the initial vertical speed, this speed is in the window

           Voy = (Y + ½ g t²) / t

           Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

            Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

             Vy = Voy - gt₂

             Vy = 0 -g t₂

             t₂ = Vy / g

             t₂ = 27.7 / 9.8

             t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

            X = Vox t₂

            D = 340 2.83

            D = 962.2 m

           

            Y = Voy₂– ½ g t₂²

            Y = 0 - ½ g t2

            H = Y = - ½ 9.8 2.83 2

            H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

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