solenoid 2cm=r 20cm=l 250turns carriesI=15A.what is magnetic flux through circular cross sectional area?
367740
You already have B.  Calculate A from the dimensions given.  This will give you the flux.  Divide that by area of a circle of radius 2 cm to get your answer.\n\nAlso, I don't know if this will help, but there's a specific equation for the magnetic field of a solenoid:  B(solenoid) = (MUo)(N1/l)(I)  [MU=the Greek letter that looks like an M, and o is a subscript zero]  That last character is a capital I and the one before that is a small L.
594911
Radius of the cylinder = 225 mm / 2 = 112.5 mm\n\nThe length is given in metres, so work in metres \n112.5 mm = 0.1125 m \n\nCross sectional area of cylinder = pi * r^2 = 0.03976078... sq m\n\nVolume = length * cross sectional area \n= 1 m * 0.03976078 sq m = 0.03976078 cu m\n1 cu m = 1000 litres\ntherefore volume = 39.8 litres to 3 s f
564373
dia = diameter, right?\n\nok...\n\nV = pi * r²h\n\nV = pi * 0.5m² * 1.2m\n\nV = pi * 0.3m\n\nV = around 0.942478m³.\n\nSo a cylinder with diameter 500 mm and length 1200mm gives us 942.478 litres of fuel.\n\nThat is though, if we ignore the thickness...\n\nEDIT\n\nWhoops, I mistook the diameter to be the radius. The correct answer is the above divided by 4, or about 235.62 litres.
954946
force required would be area under shear stress multiplied by the shear stress value(tau) of a particular material, in this case steel.\n\nthe first question is easy. the required area is the cross sectional area of that bolt. pi(r)². please substitute r in meters only.\n\nfor the second question, the shear area is not the area of the circle plate but the area of the ring around it. this area is effectively a rectangle, and can be calculated by multiplying the circle's circumference (2pi x r) by the thickness of the circle (0.0058m).
1030003
1 I don't know sorry :&#92;\n\nand 2 its simple. an electromagnet!
1083571
1) Magnetic field due to a current passing in a circular loop:\n        B=(PIN)/2r where\n  P is the permeability of the medium\n  I is the electric current\n  N is the number of turns\n  r is the radius of the loop\n\n2) Magnetic field due to a current in a soleniod:\n         B=(PIN)/L\n         B=PIn     where\n   P is the permeability of the medium\n   I is the electric current\n   N is the number of turns\n   L is the length of the coil\n   n is the number of turns per unit length\n\nAbout V & B\ni don't know any formulas that contain V & B \n\ni hope i could helped u
756487
176.53 cm\n\nhttp://www.csgnetwork.com/csgcvtftin2cmeterscalc.html
940333
54 m/s / 52 rev/s = 54/52 m/rev\nC = 2 pi r\nr = 54 / (104 * pi) = 0.165 m (rounded down)
910186
You have to get the angular velocities using conservation of angular momentum before you even THINK about energy.\n\nFor the device: I = .5mr² = .5*155*2.63² = 536.06 kg.m²\nω = .564 rev/sec * 2π rad/rev = 3.5437 rad/sec....so, \nLd = Iω = 536.06*3.5437 = 1899.64 kg.m²/sec\n\nFor the man:  I = mr² = 59.4*2.63² = 410.86 kg.m²\nω = v/r = 2.73/2.63 = 1.038 rad/sec     so,\nLm = Iω = 410.86*1.038 = 426.47 kg.m²/sec\n\nThe initial momentum is therefore 1899.64 + 426.47 = 1946.11\nThe momentum is conserved, so:\n\n1946.11 = (Id +Im)*ω, from which we can solve for the final ω:\n\n= 2.0552 rad/sec\n\nYou finish it up from here; I'm going to bed!
404470
a) AX = 2.25 cm.\n\nb) area AXY upon area ABC = 0.5625\n\nc) area AXY upon area trapezium XYCB = 0.4375
112213
area of a circle: pi*r^2\nCircumfence: 2*pi*r
1128176
work = change of kinetic enegy\nW = 1/2 m (vf² - vi²)\nW = 0.5 * (0.250 kg) * (16² - 20²) m²/s²\nW = -18 J
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
216604
I get 1276.68 mmHg.\n\nCalculate as:\n\nPressure = Force/Area.\n\nThe Area is given, 19 cm^2\n\nForce is going to be weight here. So we will use gravity as the acceleration term 9.8 m/s^2\n\nForce = 33kg * 9.8 m/s^2 = 323.4 N\n\nPressure = 323.4 N/19 cm^2 = 17.021 N/cm^2\n\nUse a conversion factor to convert N/cm^2 to mmHg and you'll get 1276 mmHg
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
754118
E1=M*g*h, M=(your mb)=32kg, - pot energy of the system at rest,\nh=2m and g=9.8m/s^2.\nE2=kinetic of M=32kg\nE3=kinetic of m=20kg\nE4=kinetic of pulley u=5kg\nE5=potential energy of m at the height h, E5=m*g*h\nThus energy conservation: E1=E2+E3+E4+E5.\nE2+E3=(M+m)*V^2/2, V to be found.\nE4=I*W^2/2, angular speed being W=V/r, r=0.1m, and I=u*r^2/2 for your pulley.\nHence 2*M*g*h=(M+m)*V^2+(u*r^2/2)*(V*r)^2+2*m*g*h,\n(M+m+u/2)*V^2=2*(M-m)*g*h, - does not depend on r.\nV=2.08m/s, don’t believe & check it! – good picture gal – a piece of art.
883859
energy dissipated in the region AB is mu*m*g*d=3.75J\nenergy stored in spring = 0.5 *k*x^2  = 0.5 m*(vb)^2 + energy dissipated in region AB   -(1)\nwhere vb is the velocity of the block at the bottom of the loop\nvb>=sqrt( 5*g*r) for completing the circle vetically\nNOTE: velocity at the top of the loop is atleast sqrt(g*r) so that the centrifugal force balances mg and the block doesnt lose contact\nfrom ---(1) : 0.5 kx^2 >= 0.5* m * 5*g*r +  7.5\nso x = 0.5357 metres