Calculate the pressure exerted on a disk in the lower back having an area of 24 cm^2 if a woman stands erect and supports the weight of her upper torso on that disk. The mass of the upper torso is 17 kg. Express your pressure in mmHg. Express your answer in three significant figures.
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Mass of 17kg, force of 170 Newtons, area of 24cm^2 (= 0.0024m^2)\n\nP = F / A\nP = 170/0.0024 = 70833.33... Pa or 70.8333 kPa.\n\n760.0 mmHg equals 101.325 kPa,\n\n70.8333kPa = 531 mmHg [3sf].
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a)\nGravitational force F between two bodies with masses m1 and m2 separated by a distance r is described by the equation\nF=G*m1*m2/r^2, where G is a constant with value 6,67*10^(-11) N*m^2.kg^2.\n\nWe know that F=34.3 N and m2 = 3.5kg and r = 7.4*10^6 m\nSubstituting we get \nm1=8.05*10^24 kg\n\nb)\n\nDensity d = m / V\nSince the planet is a sphere its volume is V=(4/3)*Pi*R^3\n\nSo d = 4740 kg/m^3
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Facts to remeber while solving the physics problem\n1. A force that causes an object with a mass of 1 kg to accelerate at 1 m/s is equivalent to 1 Newton.\nWe will also be using the principles based on Newton's Laws of motion.\nEarth attracts a mass of one kilogram with a force of about 10 newtons.\nIn Physics weight and mass are different. The mass of a body does not change, the weight will depending on the gravitational pull acting on it.\n   \nF = m x a       \nWhere    m = mass (kg) \n      F = force (N) \n      a = acceleration (m/s/s) \nTo calculate the downward pressure on the bottom of the bucket .\nWe first calculate the force on a 1 Kg bucket (given) due to 30 cm(change to metres 1 m = 100cms) of water\nF =  1 x 0.3 x 0.3 = 0.09N\nTherefore total downward force on the bottom of the bucket is the addition of the two.\nTotal downward force = 10N + 0.09 N = 10.09N\n\nTo calculate weight of full bucket.\nWe know bucket weighs 1 Kg\n12 litres of water = 12 kg\ntherefore total mass = 13 kg\n1 Kg = 10N\nTherefore 13 Kg = 130N\nWight of full bucket = 130 N
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The water pressure t a given depth is give as the weight of the water above that depth divided by the area it is being applied.\n\nSo imagine a column of water extending from the top of the lake down the depth.  Calculate the volume of this column of water then using the density of water (1 grams / mL) find the mass.  The take the mass of the water and find its weight (W = mg).  The weight it is the force exerted by the water, but pressure is the force of the water divided by the area it is applied.  So divide the force by the area of the base.\n\nEDIT:\nI see you added the density of water (even though I think my density [1 g/cc] is better).  I will leave you to do your own homework.\n[I also refuse to work with such an ugly unit, I prefer metric]
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Density = mass/volume\nDensity = 615/105 gm/cm3 = 5.86 gm/cm3
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Let Mp = planet mass.\n Ms= Sun mass. Then:\nd = distance or radius. \nF= G*Mp*Ms/d^2= Mp*a\na = centripetal acceleration = V^2/d\nyou must work with this equations and get the answer.
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To work out the moments (forces exerted), multiply to mass by the horizontal distance, so\nPalms> 68x0.7=47.6N\nFeet> 68x1.0=68N\n(Where N is Newtons, the unit of force)
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The work done is force times length. The force acting in this case is just gravitational force which is also the weight. So you will get \n\nA=M*L\n\nM-weight, L-length, A-work\n\nDon't forget to convert Length to meters.
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mass = density x volume\n\nvolume = base area x depth = 0.05 m2 x 6 m = 0.3 m3\n\nm = 1000 kg/m3 x 0.3 m3 = 300 kg\n\nweight = mass x g\nG = 300 kg x 9.8 N/kg = 2940 N\n\nPressure = density x depth x g\np = 1000 kg/m3 x 6 m x 9.8 N/kg = 58800 N/m2 or Pascal\n\nAdvice> learn the formulas!
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If we consider the volume of water is known.(If not known then use graduated cylinder and find the volume)\nMass of water can be calculated using following formula:\n\nDensity(ρ)=mass(m)/volume(V) , where\nρ is the object's density (measured in kilograms per cubic meter) \nm is the object's total mass (measured in kilograms) \nV is the object's total volume (measured in cubic meters) \n(Temp (°C)/ Density (g/cm3) \n30/ 0.9957 \n20 /0.9982 \n10 /0.9997 \n0 /0.9998 \n−10/ 0.9982 \n−20 /0.9935 \n−30/ 0.9839 )
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G = 6.67 x 10^11 N.m2.kg-2, a universal constant\nM = 5.9742 × 10^24 kilograms, the mass of the earth\nr = 6378.1 kilometers, the radius of the earth\n\nSo add 390 to r.  G times M divided by r squared is your new g.
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dude u need to start doin your own hw. ill do it this time\n5. well you have to check all the forces on the bag. the first force is gravity, which is downwards and its 9.8m where m is the mass (Force = mass x acceleration) so gravity force is 20.5kg * 9.8 = 200.9N. then the 2nd force is the force of your own movement. you accelerate it upwards at 5m/s^2. so you apply a force of 5x20.5 = 102.5N on the bag upwards. but this is the thing. it gets confusing. the force is applied on the bag, not on the items in the bag. the items in the bag  is what  makes the bag weigh so much. so if you pull the bag up with 102.5N force, the items in the bag pull downwards with the same force (another one of newtons laws) which means the total downwards force is 102.5 + 200.9 = 303.4N downwards. which is more than 245N so the bag will rip. \n\n6. well in general F  = ma. but the F is total force. in this case, there is a friction force along with the applied force.\nso Fa + Ff = ma\n   37 + Ff = 5 x 5.8\n   Ff = 29 - 37\n   Ff = -8N     its negative because its opposing the applied force. in this case i can choose which direction i want to be positive, i chose the direction that the object is moving. you could put negatives in front of the applied force, friction force and acceleration   and youll get the same answer.  so friction force is 8N.\nb. the coefficient of friction is calculated by Ff = u x Fn\nwhere Fn is the normal force and u is the coefficient. you already know Ff. the normal force is easy its basically  the force tha topposes gravity. if the object isnt moving up or down (if its on a surface). so then gravity is 5 kg x 9.8 = 49N \nso then Ff = u x 49\n    8 = u x 49\n u = 8/49\n u = 0.1632..... coefficient of friction is 0.1632\n\n7.\na.  the force is 107N and the mass is 5.1kg. so acceleration is 20.98m/s^2\nb. so acceleration = Vf-Vi/t\n  20.98 = Vf - 0 / 10s\n  Vf = 10 x 20.98 = 209.8m/s\nfinal velocity is 209.8m/s\nc. the net force that acts on the object is only the downward gravity force which is 9.8x 5.1 = 49.98.\nd. the direction of their velocity goes down when the velocity becomes 0. so you can do this by saying acceleration is gravitational and it is -9.8. initial velocity is 209.8 and final velocity is 0. and now u gotta find time. so  a = Vf-Vi/t\n-9.8 = -209.8 / t\nt = 209.8/9.8\nt = 21.4s   so it starts going down   21.4s after 10s after its been released (the first 10 s  its accelerating upwards)   so it starts going down 31.4s after its been realeased.
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F = GmM/r^2; where G is a constant, m and M are the two masses, and r is the distance between them.\n\nm + M = 4 kg; so m = 4 - M\n\nF = G(4 - M)M/(.25)^2 = 2.5*10^-10; where G = 6.67 × 10−11 N m^2 kg^−2 \n\nUse the above to solve for M, then m = 4 - M to find m.  \n\nAnd there you have it, you do the math.
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base=7cm x height=5cm/2 equals the area.\n\narea=17.5cm squared
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you divide the mass by the volume\n1.2.14 g/cm^3\n2..82 g/cm^3\ngood luck!
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The pressure will be Force/ Area\nP=1/2.5*10^-4= 400Pascal\n\nSince the obstruction is att end the area of liqid will be that of needle so Force acting hre is Pressure * area\n\nF=400*5*10^-3*10^-4 = 2*10^-4 Newton\n\nNow pressure of 50 mm of mercury is\nDensity * g\n=13600*9.8  ~= 1.36*10^5 Pascal\n\nThis must be equal to pressure in Chamber of Syrnge.\nso Pressure in chamber\n\n=F/Area\n1.36*10^5=F/2.5*10^-4 m2\n\nF=1.36*2.5*10\n\nF=34 Newton\nSo
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The question is easy enough if you think about it properly.  What you want is the speed at which the centrifugal force (given by mv^2/r) equals the gravitational force given by the bank, and that will be mg sin(2.5 degrees).  Note that m cancels out in this, and you are given r, so v is the only unknown and you can solve for it with a scientific calculator.