Answer:
unloosen the tight attire parts,open all windows for better air circulation,if he/she does not react place your hands on the chest and press gently three times per interval while giving them air through the mouth CPR then when they react place them to lie horizontal face sideways
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Hope this helps
Given the time, the final velocity and the acceleration, we can calculate the initial velocity using the kinematic equation A:

A skateboarder flies horizontally off a cement planter. After a time of 3 seconds (Δt), he lands with a final velocity (v) of −4.5 m/s. Assuming the acceleration is -9.8 m/s² (a), we can calculate the initial velocity of the skateboarder (v₀) using the kinematic equation A.

Given the time, the final velocity and the acceleration, we can calculate the initial velocity using the kinematic equation A:

Learn more: brainly.com/question/4434106
The tennis ball lands at a point 40.4 m from the base of the building.
The tennis ball is projected with a horizontal velocity <em>u</em> from a window, which is at a height <em>y</em> from the ground. The ball lands at a distance <em>x</em> from the base of the building. Let the ball take a time <em>t</em> to reach the ground. In the time <em>t</em> ,the ball falls a vertical distance <em>y</em> and also travel a horizontal distance <em>x</em>.
The initial vertical velocity of the ball is zero, since the ball is projected in the horizontal direction. The ball falls down under the action of gravitational force.
Thus, use the equation of motion,

rewrite the expression for <em>t</em> and calculate the value of <em>t</em> using 9.81 m/s²for <em>g</em> and 500 m for <em>y</em>.

The horizontal distance <em>x</em> is traveled using the constant velocity <em>u </em>since no force acts on the ball in the horizontal direction.
Therefore,

Substitute 4 m/s for <em>u</em> and 10.096 s for <em>t</em>

Thus, the ball lands at a point 40.4 m from the base of the building.
D. Travels slower in glass than in air..bends away from normal