Physics
First find the orbital speed(V)by equating force of attraction (between the two bodies) with the centrifugal force. Having found the speed you can find the period T= 2*pi*R/V
Physics
S=(1/2)gt^2 since u=0 4000=(1/2)*10*t^2 t=sqrt(800)secs - time to reach ground So hor. distance covered= 80*sqrt(800) = 2263m
physics
The force required to move the crate at constant speed should just be greater than the frictional force. F (ext.)= 180 N Work done = F*s = 180*5.5 = 990 J
Physics(Please help)
W = p*dV = p(Vf - Vi)--for isbaric p is constant. Plug in values of W,p and Vi to find Vf
Physics
let q1= -2C, q2= +5C, q3= -4C also r13=10+4=14cm=0.14m & r23= 4cm = 0.04m a) Force on q3 due to q2: F32= k*q3*q2/(r23)^2 towards left b)Force on q3 due to q1: F31= k*q3*q1/(r13)^2 towards right c) Net force on q3 = F32-F31
Physics
Assume the string makes an angle theta with the horizontal and tension in the string is T. T*Cos theta = m*v^2/r (centrifugal force) and T*Sin theta = m*g By dividing: Tan theta = g*r/v^2 = 9.8*2.2/7.2^2 = 0.41 So Theta = 22.5 deg Now T=m*g/Sin22.5 = 79.3N
Physics
Assume the rope makes an angle theta with horizontal and tension in the rope is T. Let the normal reaction at the pole and wall contact point is N (in horizontal dir.) T*Cos theta = N ....(1) T*Sin theta + mu*N = Mg ........(2) So, Tan theta = (M*g-mu*N)/N or, Tan theta = M*g/...
Physics
How much is the ramp's inclination with the horizontal?
Physics
v^2 = u^2 - 2*a*s here, v = 0 (final speed) u = 24 m/s a = 0.065 m/s^2 Solve above to obtain s - distance covered while slowing down.
Physics
Since the skier is pulled at a constant speed, the pulling force (for one skier)is given by: F = m*g[sin theta + mu_k*cos theta] = 80*10[sin25+0.15*cos25] = 446.8N (along the incline upwards) Work done by F in moving distance of 230m: W=F*x = 446.8*230 =102,776 Joules Power re...
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