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A suitcase (mass m = 18 kg) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures 0.50 m × 0.15 m. The elevator is moving upward with an acceleration of magnitude 1.6 m/s2. What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

Respuesta :

krlosdark12

Answer:

P=2736 Pa

Explanation:

According to Newton we have that:

∑[tex]F=m*a\\[/tex]

A force is exerted by the elevator to the suitcase, according to 3th Newton's law an equal force but in the opposite direction will appeared on the suitcase, that is:

∑[tex]F=m*g+m*a=m*(g+a)[/tex]

[tex]F=205.2N[/tex]

We know that the pressure is given by:

[tex]P=\frac{F}{A}\\P=\frac{205.2N}{(0.50m)*(0.15m)}\\P=2736N/m^2=2736Pa[/tex]

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