Recent content by Aerospace

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    Fluid Mechanics - Torque at hinge of closed vessel

    The closed vessel contains water with an air pressure of 10 psi at the water surface. One side of the vessel contains a spout closed by a 6-inch diameter circular gate hinged along one side. Horizontal axis of the hinge is located 10 ft below the water surface. Determine the minimum torque that...
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    Fluid Mechanics - Pressure - U-Tube Manometer

    so the force P = p1*A, correct? I stated that above in my post...is this incomplete? And in any case, if Pair was not neglected, how would you write it into the equation?
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    Fluid Mechanics - Pressure - U-Tube Manometer

    hmmm then what would it be? because my book says "at the open end the pressure is zero."
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    Fluid Mechanics - Pressure - U-Tube Manometer

    well, isn't the pressure at the open end supposed to be equal to zero?
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    Fluid Mechanics - Pressure - U-Tube Manometer

    Hmmm. I did take that into account when I did -\gamma(Hg)*h
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    Fluid Mechanics - Pressure - U-Tube Manometer

    Ok, so far this is what I've got. Let me know if I'm on the right track. If you take the pressure at the piston as P1 then, P1 + \gammaH2O*h1 - \gammaHg*h = 0 Solve for p1 and then P(Force)= p1*Area
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    Fluid Mechanics - Pressure - U-Tube Manometer

    Here's a scan of the piston I'm talking about.
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    Fluid Mechanics - Pressure - U-Tube Manometer

    A piston having a cross-sectional area of 0.07 m^2 is located in a cylinder containing water as shown in the figure. An open u-tube manometer is connected to the cylinder as shown. For h1=60 mm and h=100 mm, what is the value of the applied force, P, acting on the piston? The weight of the...
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    Non-defense Aerospace jobs?

    Hi, I am an undergrad Aerospace engineering student, and I am going to start my junior year this Spring. I'm really interested in what I'm studying, but the job market for Aerospace engineers is predominantly federal. Does anyone know of, or can think of, or work for a company that is in need...
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    Lenses and Mirrors

    yup because m = distance of image/distance of object so we can say that m = 4 hence di = 4 x do (where i and o represent image and object) Then we can plug that into the equation for focal length which is 1/di + 1/do = 1/f and we know the f, so we can solve for 'do' to get the answer :)...
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    Lenses and Mirrors

    haha...i already knew those formulas. but what's stumping me is...i only know f and the magnitude -.- is there something that i'm not seeing? i can be totally blind sometimes!
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    Lenses and Mirrors

    Stumped...lol.. A concave mirror has a focal length of 62.4 cm. Determine the object position for which the resulting image is upright and four times the size of the object. What formula do I use for this question? :confused:
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    Induced emf

    YUP! That's what I was thinking too. One of the equations for induced emf is E=Blv but how do I get the velocity from the gravity because i am not given a time period. It's probably something just basic...but I can't seem to grasp on it at this moment.
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    Induced emf

    A 5.56m long steel beam is accidentally dropped by a construction crane from a height of 3.97m. The horizontal component of the Earth's magnetic field over the region is 28.4e-6T. Acceleration of gravity is 9.8 m/s^2. What is the induced emf in the beam just before impact with the Earth...
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    Magnetic Fields

    Thank You! Thank you so much. It still took me a while to get it, but I finally got it!! Thanks to both of you :)
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