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natka813 [3]
3 years ago
6

A hockey puck slides off the edge of a horizontal platform with an initial velocity of 20 m/s horizontally and experiences no si

gnificant air resistance. The height of the platform above the ground is 2.0 m. What is the magnitude of the vertical component of the velocity of the puck just before it hits the ground
Physics
1 answer:
stepladder [879]3 years ago
7 0

Answer:

v_{yf}=6.26m/s

Explanation:

We have to focus only on the vertical direction: the puck falls 2m with a null vertical initial velocity. We use the equation:

v_f^2=v_i^2+2ad

which for the vertical direction is:

v_{yf}^2=v_{yi}^2+2a_y\Delta y

or:

v_{yf}=\sqrt{v_{yi}^2+2a_y\Delta y}

Taking the downward direction as positive (which means a positive gravitational acceleration and a positive displaceent) we have for our values:

v_{yf}=\sqrt{(0m/s)^2+2(9.8m/s^2)(2m)}=6.26m/s

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KI + Cl2 ---&gt; KCl + I2<br> Balance the single replacement chemical reaction.
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2KI   +  Cl₂   →   2KCl     +    I₂

Explanation:

The reaction equation is given as:

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The problem at hand is to balance this chemical reaction. To solve this problem we use a mathematical approach;

        aKI   +  bCl₂   →    cKCl     +    dI₂

 Conserving K : a = c

                     I :  a  = 2d

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 Now let a = 1, c  = 1 , d  = \frac{1}{2}, b   = \frac{1}{2}, ;

   Multiply through by 2;

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an average net force of 31.6 N is used to accelerate a 15.0 kg object uniformly from rest to 10.0 m/s leftward. What is the chan
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Answer:

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