Simple Fluid pressure-force question

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The discussion centers on calculating the net force on a rectangular window due to differing air pressures inside and outside a house. The window dimensions are 3.20 m by 2.40 m, with outside pressure at 0.948 atm and inside pressure at 1.00 atm. The correct area calculation for the window is needed, as the user initially misapplies the area formula. The pressure difference is calculated using atmospheric pressure set at 101 kPa. The user expresses confusion about their approach but ultimately decides they no longer need assistance.
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Homework Statement



A rectangular window in a house has dimensions 3.20 m x 2.40 m. As a storm passes by, the air pressure outside drops to 0.948 atm while the pressure inside the house remains 1.00 atm. What is the magnitude of the net force pushing on the window?
Take atmospheric pressure to be 101 kPa

Homework Equations


P= F/A
Area of rectangle= 2(length + width)
1 atm = 101 kPa


The Attempt at a Solution



A = 2(3.2+2.4)= 11.2 m^3
P= 101 kPa (it only requires pressure inside house because that's the force of interest)

p=f/a
p*a=f
11.2* 101 kPa

what am i doing wrong? is it the units?:confused:
 
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grewas8 said:
Area of rectangle= 2(length + width)

That is really wrong and you should know better.
 
nvm
dun need help
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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