I really need help with a physics problem?
vx = v0*cos(53 deg); vy0 = v0*sin(53 deg); and we need -15 = vy0*t - (1/2)gt^2, while 40 = vx*t. (Consider the "v"s to be given in m/s.) -15 = v0*sin(53)*t - (4.9)t^2 and 40 = v0*cos(53)*t. Then v0 = 40/[t*cos(53)], so -15 = 40*tan(53) - (4.9)t^2 => (4.9)t^2 = 68.08 => t = 3.727 => v0 = 17.8 m/s. (part B) If v0 = 8.9 m/s, then vy0 = 7.108 m/s and vx = 5.356 m/s. The time to cross to the other building would be 7.468 seconds, but the time to fall to the ground would be found from -100 = (7.108) t - 4.9 t^2 => 49 t^2 - 71.08 t - 1000 = 0 => t = 71.08/98 + + (1/98)*sqrt(71.08^2 + 4000*49) = 5.30 seconds. So the motorbike would land on the ground, about (40 m)(5.3/7.47) = about 28.4 m away from the base of the first building.
Can you help me solve this physics problem involving an uncharged RC circuit?
The charge on a capacitor q = C*v where v = v0*(1- e^-t/τ) Here t/τ = 2.0 So q = C*v0*(1- e^-2) = 15x10^-6*25*(1 - e^-2) = 3.24x10^-4C
Physics problem help please?
1) 85km/hr * 1m/s / 3.6km/hr = 23.6 m/s kinematics of the cars: cop: s = 23.6m/s * t + ½ * 3m/s² * (t - 2)² speeder: S = Vo * t "after accelerating for 6.00s" means t-2 = 6s, so t = 8s Then s = S when 23.6 * 8 + 1.5 * 6² = Vo * 8 242.9 / 8 = Vo = 30.4 m/s = 109 km/hr
How do I calculate stopping distance for problems in physics?
1) Acceleration = (f/m) = 1400/51, = -27.451m/sec^2. Initial V at water entry = sqrt.(2ah) = sqrt.(54.902 x 4.4) = 15.542m/sec. Drop height from board to water = (v^2/2g) = 12.3242m. metres. Total distance = (12.3242 + 4.4) = 16.724 metres. 2) Height from which it starts = (sin 30) x 2.92 = 1.46 metres. GPE = (mgh) = 8.8 x 9.8 x 1.46, = 125.91 Joules. This becomes KE at bottom of ramp. V at bottom = sqrt.(2KE/m) = 5.3494m/sec. Acceleration = (v^2/2d) = -2.611m/sec^2. Friction force = (ma) = 22.9768N. Normal force = (8.8 x 9.8) = 86.24N. a) Coefficient = (22.9768/86.24) = 0.266. b) Change in KE due to friction = 125.91 Joules (above).
Please help me solve the last part of this physics problem!!?
d) Height h = 2 m Translational speed at the bottom of the hill v0 = 10 m/s Moment of inertial I = 1/2 mr^2 Angular velocity w = v/r Let translational speed at the top of the hill = v Translation kinetic energy = 1/2 mv^2 Rotational kinetic energy = 1/2 I w^2 = 1/2 * (1/2 mr^2) * (v/r)^2 = 1/4 mv^2 Final kinetic energy = 1/2 mv^2 + 1/4 mv^2 = 3/4 mv^2 Initial kinetic energy = 3/4 mv0^2 Change in kinetic energy = 3/4 mv^2 - 3/4 mv0^2 Work done by gravity = -mgh From work energy theorem, 3/4 mv^2 - 3/4 mv0^2 = -mgh Dividing by m, 3/4 v^2 - 3/4 v0^2 = -gh Multiplying by 4/3, v^2 - v0^2 = -4/3 gh Or v^2 = v0^2 - 4/3 gh = 10^2 - 4/3 * 9.8 * 2 = 100 - 26.13 = 73.87 Or v = sqrt(73.87) = 8.6 m/s Rotational speed = v/r = 8.6/0.7 = 12.3 rad/s Ans: 8.6 m/s 12.3 rad/s
Help with physics (angular speeds and such).?
Let's start with things we know. Ball has mass M. Its velocity at time t is given as v(t) = Vo + at where a is the acceleration (negative) due to friction, or a = -µg. Then v(t) = 6m/s - 0.19*9.8m/s² * t = 6 - 1.862t Torque τ = Iα = (2M(0.11m)²/5)α = 0.00484Mα But also τ = F * r = µMgr = 0.19 * M * 9.8m/s² * 0.11m = 0.205M so α = 0.205M / 0.00484M = 42.3 rad/s² angular velocity ω(t) = ωo + αt = 0 + 42.3rad/s² * t The ball stops sliding when v(t) = ω(t) * r 6 - 1.862t = 42.3*0.11*t = 4.655t 6.517t = 6 t = 0.92 s (b) s = Vo*t + ½at² = 6m/s * 0.92s - ½ * 1.862m/s² * (0.92s)² = 4.7 m (c) v = Vo + at = 6m/s - 1.862m/s² * 0.92s = 4.3 m/s
Physics Problem: Energy Conservation?
I can't figure out this problem An empty crate is given an initial push down a ramp, starting with a speed v0, and reaches the bottom with a speed v and kinetic energy K. Some books are now placed in the crate, so that the total mass is quadrupled. The coefficient of kinetic friction is constant and air resistance is negligible. Starting again with v0 at the top of the ramp, what are the speed and kinetic energy at the bottom? So far I got that energy at the top would equal energy at the bottom (conservation). So fo only the crate: KE0= 1/2mv0^2 (Vo=sqrt(2KE0/m)) KEf=1/2mv^2 then for the books and the crate (mass=4m) Ko=1/2(4m)vo^2 Ko=2m(2k0/m)=4k0 so because top and bottom kinetic energies are equal, would the kinetic energy at the bottom just be 4K? And speed of the 4m mass would be sqrt(8k/m)? Where does potential energy come into play? Please help! Thanks.
Physics homework problem - Related to Doppler Effect?
I have a homework problem that I am completely stuck on, I've exhausted all my thoughts and keep getting the wrong answer. Here is the problem: A submarine traveling at 12 m/s toward an aircraft carrier emits a 2270 Hz sonar pulse. The reflected pulse returns with a frequency of 2295 Hz. What is the speed of the aircraft carrier? (Positive or negativemeans the carrier is moving toward or away from the submarine, respectively.) The speed of sound in water is 1500 m/s. Answer in units of m/s. ___ The way I set it up was as follows: f '=[(v-v0)/(v - vs)]*f With f ' representing the frequency of the observer, v representing the speed of the wave, v0 representing the speed of the observer, vs representing the speed of the source, and f representing the frequency of the source. So first was the boat as the observer and the sub as the source. So its setup is: f '(boat) = [(1500m/s-0(wall)) / (1500m/s - 12m/s)]*2270 Hz f '(boat) = observed frequency by boat = 2288.306 Hz Then I use that value as the source and I repeat the formula with the observer now being the sub: f "(sub)= 2295 = [(1500m/s - 12m/s) / (1500m/s - vs(boat)]*2288.306 and solved for vs(boat) I came up with 16.3399 m/s.....and its wrong. I don't know where I went wrong. Anyone?