Power = force time velocity= Volt divided by resistance
215571
The Watt is a unit of Power.\n\n(I'm pretty sure P does not equal vf though)\n\nEDIT - No you're right. I had to work it out for my own peace of mind though!\n\np=e/t\nw(e)=fd\np=fd/t\np=f(d/t)\np=fv
572706
Power = work done per unit of time.\n\nP=W/T
380026
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
509118
= 9u-10u + 6v = 6v - u
383811
a=(delta V)/(delta t)\n\ndelta V = V_final - V_initial = 0 - 3.2\ndelta t = 0.01 sec\n\na=3.2/0.01=320 m/s2\n\nforce=mxa=0.084x320=26.88 N
440292
v = rω\n=10 * 7π\n≈ 219.9 ft/s
145722
power is the amount of work done in unit time \nit is measured in Joules per second , dude
509087
The ball\nVt= Vo + at\nVo= 0 hence Vt= at\nF = m * a = 0.350 * a\na= F/0.350\n28 = F/0.350 * t\nF= 28 * 0.350 /t  = 9.8 /t\n\nThe man\nt man = t ball\nF man = F ball ( Newton's law)\nF = 9.8/t\nF = m * a = 65 * a\n9.8/t = 65 * a\na = 9.8/(65*t)\nV=a*t\nV= 9.8/(65*t) * t = 9.8/65 = 0.151 m/sec
420968
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
398368
P = FD/t = 300*9.8*5/8 = 1837.5 Nm/sec = 1.8375 kw
64315
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
298394
If the equation is x=vt, then v is in m/sec.
500885
In a purely resistive circuit, the power factor (PF) is 1 and you *can't* adjust it.\n\nAnd, since this is the 3'rd time you've asked the question and gotten the same answer from one other person and myself, do you even know what 'Power Factor' *is*?\n\n\nDoug
121182
Use the node voltage method. Label each node with a voltage, then apply Kirchoff's current law at as many nodes as you need to generate enough equations to solve for any unknowns you have introduced.\n\nLet the node at the bottom of the circuit be ground, and then try to figure out the voltages at the other nodes relative to ground. Clearly the voltage at the node on the top will be 14V (I'll just call this V1) but no other voltages are apparent. So call the voltage at the node between R1, R3, and R4 Vx and the voltage at the node between R2, R3, and R5 Vy.\n\nNow that we have labeled all the node voltages, you can figure out every single current in the circuit in terms of V1, Vx, and Vy (using Ohms Law, the voltage drop across a resistor is equal to the resistance times the current in the direction of the voltage drop). For example, we know the current going down through R1 is (V1-Vx)/R1, the current going right through R3 is (Vx-Vy)/R3, etc. Now we just have to solve for the unknowns Vx and Vy.\n\nKirchoff's current law (KCL) says that the sum of the currents leaving a node is equal to zero. If we apply this law to the node with voltage Vx and the node with voltage Vy, we'll get a pair of equations that we can use to solve for Vx and Vy.\n\nApplying KCL at the node with voltage Vx should give you the following equation:\n\n(Vx-V1)/R1 + (Vx-Vy)/R3 + Vx/R4 = 0\n\nThe first term is the current going up through R1 (we already figured out the current going down through R1, but we want currents that are leaving node Vx so we had to flip the sign), the second term is the current going right through R3, and the final term is the current going down through R4.\n\nI'll leave it to you to figure out the equation for node Vy. Once you have that you'll have two equations and two unknowns and you can solve for everything. Once you have Vx and Vy, you can easily figure out a numerical value for any current in the circuit.\n\nBTW, the node-voltage method is pretty powerful and can solve most linear circuits. Check out the link below, it was written by an old EE prof of mine who is very thorough.
157666
The total resistance is 12 + 8 + 6 = 26 ohms. In series the total resistance is the sum of individual resistors. Apply ohm's law V=IR, I = 20/26 Amperes. The voltage drop across each resistor is found by multiplying the current with the resistance of each resistor.
92914
Starting with Ohm's law:   P = I * V\n\nSo in your example: \n P1 = 120V x 25A = 3000W\n P2 = 120V x 30A = 3600W\n\n Ptotal= P1 + P2 = 7600W or 7.6kW\n\nNow, energy consumed in kW-hr is just the power multiplied by the duration in hours:\n\n  Say that you held this load for 2 hours, then:\n  \n  E = P x t = 7.6kW x 2 hours = 15.2kW-hr
344253
Hello, I would be glad to help you with this:\n\n a=(v-c)/t\nat=v-c\n-c=at-v\nc=v-at \n\n I hope this helps and good luck with the rest of your course work ,Please Please remember me for best answer