Physics problem help needed (first correct answer gets best answer)?
Emf induced = -Bbv where B is the A) What is the magnitude of the induced current? B) Later the loop is fully in the field, still moving at 29 cm/s. What is the magnitude of the induced current? C) Eventually the loop begins to emerge from the field, still at 29 cm/s. What is the magnitude of the induced current? B =0.85 T L = 12 cm b = 8cm A) Emf induced = -Bbv where B is the magnetic induction, b is the breadth and v is the velocity of the loop. Current = Emf/ resistance = -Bbv/R = - 0.85*0.08*0.29 /3.5 = 0.0056 A ======================= B) The flux is cut at the rate of 0.29m/s . The same current persists. C) The flux is cut at the rate of 0.29m/s. The same current persists. =============================== Another way to answer B and C , there is no variable in the formula for the current which will alter the current. B,b,v&R are all constants. =========================
PHYSICS HELP. BEST RESPONSE TO FIRST CORRECT ANSWER?
d = 0.5at^2 v = at 360 m/s = at a = 360 / t d = 0.5at^2 0.750 m = 0.5 * 360/t * t^2 (a t cancels) 0.75 m = 0.5 360 * t t = 4.17 * 10^-4 s a = 360 / 0.000417 a = 863,309.35 m/s^2 F = MA F = 0.025 kg * 863,309.35 F = 21,582.73 N
Physics 1st correct answer gets best answer?
A rigid rod of mass 5.10 kg and length of 3.30 m rotates in a vertical (x,y) plane about a frictionless pivot through its center. Particles m1 (mass=7.10 kg) and m2 (mass=1.60 kg) are attached at the ends of the rod. Determine the size of the angular acceleration of the system when the rod makes an angle of 39.5o with the horizontal
Physics question, hard! First right will get best answer!?
Let me try. The PE of the rocket at launch is = -GMm/r (where G is grav constant, M is mass of Earth, r is radius of Earth at launch point, m is mass of rocket) The KE of the rocket at launch is = 0.5 m.v1^2 (where m is mass of rocket, v1 is initial velocity = 1.8 x 10^4) So the total energy of the rocket at launch is = 0.5m.v1^2 - GMm/r This total energy must be conserved. So when the rocket is so far away that its gravitational PE can be ignored, then its remaining kinetic energy must be the same as its total launch energy. 0.5 m.v2^2 = 0.5mv1^2 - GMm/r (where v2 is its final velocity) simplifying, v2 = SQRT( v1^2 - 2GM/r ) Note that v1^2 > 2GM/r to escape the Earth's gravity
PHYSICS PROBLEM ON FORCES...HELP! First good answer gets 10pts!?
So I don't understand this at all...please explain how you solved this problem! On a slippery (zero friction) level surface, two people are pulling on a 35Kg pumpkin. One person is pulling with a force of 125N toward the east, The other person is pulling with a force of 75N toward the west. If the initial velocity of the pumpkin was 3m/s toward the west, what is its final velocity after it has slid 2.0m?
Physics riddle, first answer gets Best Answer!?
Ok, i assume you mean a force of F = F_0*sin(ω*T) with F_0 = 1 [N] and ω = 1 [1/s] to get correct units ;-) By using the symbol ω i imply, that i assume that the argument of the sine will be in radians. This means, that the sine will have completed more than a full oscillation (T > 2π). The only contribution of the sine is that after a full oscillation, because there is no work done (area under sine from 0 to π equals area under sine from π to 2π, since we need oriented areas, they cancel out). So we´ll only need the sine from 2π to 10, or, because the sine is 2π-periodic, from 0 to 10 - 2π. Now, from Newton´s second law, we know, that F = d/dt (m*v). since the mass does not change, this reduces to: F = m * dv/dt . Now we can seperate the variables to get: F * dt = m * dv dv = F/m * dt We can integrate this equation to get the velocity. The lefthandside will be integrated from 0 to 10-2π, because the particle would also be at rest at 2π, the right hand side will be integrated from 0 to v, because at t = 2π, the particle will have no velocity. ∫dv = ∫ F/m * dt v = F_0/m ∫ sin(ωt) dt v = 1/m ∫ sin(t) dt v = 1/m [- cos(t)] from 0 to 10-2π v = -1/m * (cos(10 - 2π) - cos(0)) v = 1/m * (1 - cos(3.716...)) v = 1/m * (1 +0.839... ) v = 1.839.../3.17 [N*s/kg // the s coming from the integration, N from F_0] v ≈ 0.580 [kg * m/s^2 * s/kg] v ≈ 0.580 [m/s] And that´s it :-) Hope i typed in the correct numbers into the calculator. #################################### €: Oops, didn´t read carefully enough, sorry... Didn´t have much sleep last night. Ok, so what changes is, that ω = 1°/s and that we need to calculate the complete integrals. ∫dv = ∫ F/m * dt 0 to v on the left hand side and 0 to 10 on the right hand side. v = F_0/m ∫ sin(ωt) dt v = 1/m ∫ sin(t) dt v = 1/m [- cos(t)] // from 0 to 10 s v = 1/m * (- cos(10°) + cos(0°)) v = 1/m * (1 - cos(10°)) v = 1/3.17 * (1 - 0.98...) [1/kg * N * s] same argumentation as above v = 0.015.../3.17 [1/kg * kg*m/s^2 * s] v ≈ 4.793 * 10^(-3) m/s or roughly v ≈ 0.005 m/s
Physics help! Drag force question?
drag force increases as the speed increases, so we know that the last three options cannot be correct however, unless we know more, we cannot distinguish between the first two options; for some speeds and shapes of objects, the drag force varies as the square of speed, and if this is the case, the drag force increases by a factor of 4 for slower speeds, the drag force varies with the first power of the speed, so the drag force would increase by a factor of 2 however, with no further information, we cannot distinguish between these two possible options
PHYSICS HELP...MOMENTUM!!!!!!!!!!...1S... CORRECT ANSWER GETS 10 PTS?
Assume 'left' is negative and 'right' is positive: i=initial f=final m=mass v=velocity The initial total momentum of the system is equal to the final total momentum of the system: m(1i)v(1i)+m(2i)v(2i) = m(1f)v(1f)+m(2f)v(2f) (0.205)(1.51) + (0.294)(0.401) = (0.205)v(1f) + (0.294)v(2f) 0.30955 - 0.117894 = (0.205)(-v(1f)) + (0.294)v(2f) 0.191656 = (0.205)(-v(1f)0 + (0.294)v(2f) 0.191656/0.205 = (0.205)(-v(1f))/(0.205) + (0.294)v(2f)/(0.205) 0.935 = (-v(1f)) + 1.434v(2f) 0.191656 = (0.205)(-v(1f)0 + (0.294)v(2f) 0.191656/0.294 = (0.205)(-v(1f)/(0.294) + (0.294)v(2f)/(0.294) 0.652 = -0.697v(1f) + v(2f) v(1f) = 1.434v(2f) - 0.935 v(2f) = 0.652 + 0.697v(1f) The final velocities can only be expressed in relation to each other unless one of the velocities has been provided to you.
Physics problem....I need help getting the correct answers?
A 0.125 kg frozen hamburger patty has two forces acting on it that determine its horizontal motion. A 2.30 N force pushes it to the left, and a 0.800 N force pushes it to the right. (a) Taking right to be positive, what is the net force acting on it? (b) What is its acceleration? I got -1.5 for A and B 12 but it's wrong, need help..
Help needed on physics problem! First correct gets best answer!?
(a) F = Σ GmM / d² where G = 6.674e−11N·m²/kg² The two left masses push rightward on the 2 kg mass with a force Fx = G * 2kg * (1/0.36² + cos(arctan(0.14/0.36))*4/(0.36² + 0.14²))kg/m² Fx = 4.365e-9 N The two bottom masses push up on the 2 kg mass with a force Fy = G * 2kg * (3/0.14² + sin(arctan(0.14/0.36))*4/(0.36² + 0.14²))kg/m² Fy = 2.173e-8 N F = √(Fx² + Fy²) = 2.2e-8 N ◄ magnitude Θ = arctan(2.173e-8 / 4.365e-9) = 79º ccw from +x axis ◄ direction (b) Double the dimensions, quarter the force. Therefore • The magnitude of the force will be reduced by a factor of four. • The direction will remain unchanged. Hope this helps!