Orbital Velocity Question?
112212
First solve for the mass of the planet M\n\nM = 5200 kg/m^3 x 4/3 x 3.14 x (1600 x 10^3 m )^3\n  = 5200 x 4/3 x 3.14 x 4.10 x 10^18 kg\n  = 8.93 x 10^22 kg\n\nNext, use the formula for centripetal force, and set it equal to the gravitation force between the object and the planet.\n\nG (Mm)/r^2 = m v^2/r , where r is the total radius of the object, that is 1600 km  + 630 km = 2230 km.\n\nDividing out the common factors of m and 1/r, we have\n\nG M /r = v^2\n\n(6.67 x 10^-11)N m^2/kg^2 (8.93 x 10^22 kg) / (2.230 x 10^6 m)\n\n2.67 x 10^6 N m /kg = v^2\n\n1.63 x 10^3 m/s = v\n\nCheck my math to make sure, but the physics is correct.
382082
use d=V1*t + [(a*t*t)/2]\nd=distance=490m\nV1=initial velocity = 0m/s\nt=time = ?\na=acceleration due to gravity = 9.81m/s*s\n\nusing equation above, t will come up to be:\nt=(square root)(2*d/a)\nt=(square root)(2*490m / 9.81m/s*s)\nt=9.99 sec or 10 sec.
289592
You can't-you dont have enough information.  You need to know the mass of the planet as well, to know the gravity.  Orbital velocity is determined based on calculating the gravity based on the mass, assuming all the gravity originates in the exact center of the planet.  Then, if you know how far the surface is from the center (the radius of the planet) you can calculate what the gravity is at this point...gravity decreases as the square of the distace from the center of gravity.  Now, you know the gravity at the surface, and can calculate escape velocity.
726547
Find the horizontal and vertical components of its velocity.  (Hint: one is v*sin(ø) and the other is v*cos(ø).)   Each velocity times the ariplane's mass is that direction's component of momentum.
844762
There are jthree  formulae cncerning velocity.These are v=u+at , V^2=u^2+2as ,and s=ut +(1/2) at^2 ,where u= initial velocity,v=final velocity ,a=accelaration.,s= distance travelled and t= time.Now how to decide which formula is to be used?in the problem,just see what they want you to find out and note down what is are already given in the problem, if they give you u,a and t and ask you to find out v,apply formula no.1.But if they gdonot give you time but give the distance or displacement and asks you to find v,donot hesitate to apply formula no,2 and so on.if you have any difficulty to understand.please do not hesitate to write to me through yahoo answersand i shall be glad to assist you
448264
Given v_i = initial velocity = 0; v_f = final velocity = 44 m/s; a = 9.8 m/s^2:\n\ndv/dt = a = 9.8 m/s^2\ndv = 9.8 dt\nint(dv, v_0, v_f) = int(9.8 dt, 0, t)\nv_f - v_0 = 9.8t\n44 - 0 = 9.8t\nt = 44/9.8 = 440/98 = 220/49\n\nint(dv, v_0, v) = int(9.8dt, 0, t)\nds/dt = v = 9.8t\nds = 9.8tdt\nint(ds, 0, h) = int(9.8tdt, 0, 220/49)\nh = 4.9(220/49)^2/2 = (49*22*220)/(49*49*2) = (11*220)/49 = (2200 + 220)/49 = 2420/49 ~ 49.3878\n\nAlternatively, since accelration was constant, you could always get it more quickly with the formulas that were derived with integrals as above:\n\nv_f = v_0 + at\ns = v_0t + 1/2at^2\n\nEither way you should get 2420/49 m if the initial conditions are correct.  I don't know where 122/5 comes from although it seems to be approximately a half of this answer.
987748
1.Compute initial velocity V0\nUsing equation S=V0*t-(g*t^2)/2 we find that the initial velocity V0=35/3 + 3g/2\n2.What ever goes up must come down V0*t=-(g*t^2)/2 t=2*V0/g\n3.Total time travel = 2t \nYou to the math\n\nBest\n\nHint:  the arrow will stop midair in about 5.3 seconds
752738
The 120N force on the astronaut accelerates him/her at a rate of 120/79 m/s^2. In 0.42 s, this would accelerate the astronaut to a velocity of 0.42 (120/79) m/s.\n\nSimilarly, the satellite accelerates in the opposite direction to a velocity of 0.42 (120/660) m/s\n\nThe relative velocity of the astronaut and the satellite is \n0.42 (120) (1/79 + 1/660) m/s\n\nMultiply this by 84 seconds (which is equivalent to 1.4 min), and you'll have the requested distance. \nNote: This is the distance the astronaut will be from the satellite IN ADDITION TO the length of the astronaut's arms. We started counting the time when the astronaut had straightened his/her arms.
217864
J=Iw, w=v/r (v = tangental velocity) I is diffrent for each object, or group of objects.  so J=Iv/r I=Jr/v\n\nI for uniform objects does = sigma MR^2.
666681
1/2 * mass * velocity^2.
1146651
You're actually right.  Furthermore, escape velocity is a function of distance from the center of mass (of the Earth in this case)\n\nEscape velocity applies to ballistic motion.\n\nV(esc.)=(2G*M/r)^1/2\n\nWhere G is the gravitational constant\nM is the mass of Earth (in this case)\nAnd r is the initial radius from Earth's center from which you are computing escape velocity.
162313
Let V = speed (the magnitude of the initial velocity)\na = angle, t = time, \nVx = horizontal (x) component of velocity\nVy = vertical (y) component of velocity\nAx = acceleration in x direction\nAy = acceleration in y direction\nDx = net distance traveled in x direction\nDy = net distance traveled in y direction\n\nVx = Vcos(a)\nVy = Vsin(a)\n\n  Dx = 1/2 Ax*t^2 + Vx*t\nbut Ax = 0, so Dx = Vx*t, t = Dx/Vx\n\n  Dy = 1/2 Ay*t^2 + Vy*t\n(Ay = -g, the acceleration of gravity, negative for downward)\n -1/2*g*Dx^2/(Vcos(a))^2 + Vsin(a)*Dx/(Vcos(a)) = Dy\nMultiply by V^2, and collect terms with V^2 on the left:\n(Dx*tan(a)-Dy) V^2 = 1/2*g*(Dx)^2/(cos(a))^2\nV = sqrt(1/2*g*(Dx)^2/(cos(a))^2/(Dx*tan(a)-Dy))\nsubstitute a = 30 degrees, Dx = 6.05 m, Dy = 3.05-2.2 = 0.85 m, g = 9.8 m/s^2\nV = 9.51 m/s\nFor a = 60 degrees,\nV = 8.63 m/s
847656
The mass is moving parallel to the x-axis with a velocity of magnitude, say v. then its velocity vector is V(vector) = vi,i is the unit vector along x-axis. Its momentum vector is p(vector) = m*V(vector).\n\nAt any instant let the mass be at A(x,y) so that OA(vector) = r(vector) [ O is the origin].\nthen by definition, the angular momentum,\n\nJ(vector) = r(vector) X p(vector) [ X implies cross product)\nor, J(vector) = r(vector) X m*v(vector)\nor, J(vector) = m( xi + yj) X vi\nor, J(vector) = mvy ( -k) [ since iXi=0 & jXi= -k]\n\nsince m,v,y are constants,[y is const because the particle is moving parallel to x-axis so its ordinate will remain const.] the angular momentum J will remain constant & its magnitude at a pt (x,y)is mvy ; direction negative z- axis [v= the velocity of the mass] .(ans)
810893
Oscar. You have written too much. This 125m tower stuff does not matter compared to r=6.38*10^6. Why is this talk about 1.00 km and angle theta? See what “zipper” above says. Besides I answer your last question << Why is the trajectory of the cannonball not parabolic?>>. Why should it be if both you and Newton demand it to be circular? Assume that V1 of cannonball is the 1st cosmic speed corresponding to a circular track. Let V2 be > than V1, then you receive an oval track. Take V3 > than V2, then the track may become parabolic, so that cannonball would never return back to the cannon to linger out among planets and stars. If you are intrigued calculate a set of velocities and corresponding orbits of the cannonball. If cannon is not available consider a football! I like your inventive wits – to be honest!
168503
Finally, the gravitational pull between the Earth and Sun causes the Earth to travel around, or "orbit", the Sun at a velocity of 29.8 km/sec. At the same time, the Earth also turns on its axis causing the daily cycle of day and night. This "rotational velocity" is approximately .47 km/sec. That means that at the same time we're hurdling through space at nearly 67,000 mph, we're also spinning around in circles at over 1000 mph! Whew! Almost makes you dizzy, doesn't it?
809913
1. Momentum=Massxvelocity=3600\n2.3600=.2KgxVelocity=18000\nThats all i know
758142
a) There is an equation derived from the motion in x and y that results like:\n\ny = (tanθ )x - g x²/ [2(Vo cosθ)² ] we know y, x and θ we need to solve for Vo:\n\nVo =  (x / cosθ) √[ g / (2 (x tanθ  - y )]\nIf we put  the plane at the initial hit at 1 m, y is 21 m -1m = 20 m\nAnd as we know that x = 130 m and θ= 35°\nSubstituting:\n\nVo = 41.7 m/s <<<<<<<<\n\nb) t = x / ( Vo cosθ )  Substituting: t = 3.81 s <<<<<<<\n\nc) Vfy = Vo senθ - gt   (Vfy is the final velocity of y after the time t)\nSubstituting, its components are:\nVfy = -13.46 m/s      <<<\nVox = Vfx = Vo cosθ = 34.16 m/s   <<<\n\nand its magnitude:\n\nVf = √(Vfx² + Vfy²) = 36.71 m/s <<<<<