Physics
T = I a where T = torque, I = moment of inertia (= 2.5 kg m^2), a = angular acceleration (= 3 rad/s^2) The answer will have units of N m
Physics
k = 400 N/m x = 0.08 m F = k x w = k x m g = k x m = kx/g g = 9.8 m/s^2
Physics
Just as an aside: A car (at least, a car as we understand it, with four wheels rotating on the ground) would not be able to accelerate on a frictionless track. The force providing the acceleration is the frictional force between the tires and the track. Some sort of "air ...
Physics
work = change in kinetic energy friction force X distance = change in KE -mu_k n x = 0 - 1/2 m v^2 [Remember, friction does negative work) mu_k m g x = 1/2 m v^2 Simplifying: x = v^2/(2 mu_k g) = 17.24 m Notice that in this solution, the mass of the car is not needed.
Physics
Because the charge on the proton is positive, the direction of the acceleration is the same as the direction of the electric field, the positive x direction. v^2 = v_0^2 + 2 a x (one of the equations of motion) Solving for a, and using v_0 = 0 a = v^2/2x This is the source of ...
physics
work = force (in direction of motion) X distance W = (F cos t) x cos t = W/Fx t = arccos(W/Fx) = arccos [8/(14)(0.63)] = arccos 0.907 = 24.9 = 25 degrees
Physics
I don't know if you've learned this equation: F = m r w^2 where F=centripetal force, w = angular velocity Assuming the force is in Newtons (you don't show the units), be sure to covert r = 10 cm to 0.1 m before solving.
physics
net change in energy = 0 (change in PE) + (change in KE) = 0 m g (h - H) + 1/2 m (v^2 - V^2) = 0 where h = 35 m, H = 70 m, V = 38 m/s Solve for v: [Notice that the mass, m, cancels in the equation]
physics
P = F v = w v = (800+600)(4) = 5600 W
physics
It looks like you have a good idea of how to solve this; unfortunately, your notation makes it very difficult to follow. Once you find the normal force (n), you will be working only along the x-axis, so just drop any x-subscripts, so the acceleration is simply a. Let's let...
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