A child wants to pump upa bicycle tire so that its pressure is 1.2 x 105 Pa abov

Question

A child wants to pump upa bicycle tire so that its pressure is 1.2 x 105
Pa above that of atmospheric pressure. If the child uses a pumpwith a
circular piston 0.035 m in diameter, what force must the childexert? A)54N B)76N C)120N D)89N E)240N
Pirates treasure chest lies 20.0 m below the surface of the ocean.What is the magnitude of the force that acts on the rectangular topthat is 0.750 mx 0.425 m? A.)2.39x10**3N B.)2.00x10**6N C.)9.48x10**4N D.)980N E) 4.71x10**5N A child wants to pump upa bicycle tire so that its pressure is 1.2 x 105
Pa above that of atmospheric pressure. If the child uses a pumpwith a
circular piston 0.035 m in diameter, what force must the childexert? A)54N B)76N C)120N D)89N E)240N
Pirates treasure chest lies 20.0 m below the surface of the ocean.What is the magnitude of the force that acts on the rectangular topthat is 0.750 mx 0.425 m? A.)2.39x10**3N B.)2.00x10**6N C.)9.48x10**4N D.)980N E) 4.71x10**5N

Explanation / Answer

1)The pressure inside the bicycle tire is P = 1.2 *105 Pa The diameter of the circular piston is d = 0.035 m The area of the circular piston is A = *(d2/4) The force exerted by the child is F = P * A = P * * (d2/4) 2)The height below the surface of the ocean where Piratestreasure chest lies is h = 20.0 m The area of the rectangular top is A = 0.750 m * 0.425m The magnitude of the force that acts on the rectangular topis F = P * A where P is the pressure exerted on Pirates treasurechest or F = gh * A ----------(1) where = 1000 kg/m3 is the density of waterand g = 9.8 m/s2 Substituting the values in equation (1) we get the magnitudeof the force that acts on the rectangular top. Substituting the values in equation (1) we get the magnitudeof the force that acts on the rectangular top.

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