# Posts by CANTIUS

Total # Posts: 21

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...

physics
You simply calculated T1x/T1y, you didn't take the arctan of this number. It's difficult to explain, especially without a diagram, but ignore the negative sign: theta = arctan(641.5/299.1) = arctan(2.14) = 65 degrees This angle is relative to the wall. Relative to the ...

physics
You solved it! However, the problem is not asking for the mass, but rather the weight w = 6664.6 N

physics
m = 29 kg v = 7.9 m/s

Physics
In my opinion, this is a poorly worded question. My answer is: Any object sitting on a flat surface. The question involves forces, which are vectors, which have direction as well as magnitude. So for my object, the normal force points straight up, the force of gravity points ...

physics
Since electrostatic force is inversely proportional to the square of the separation, if the new distance is one-ninth of the first distance, the new force will be 9^2 = 81 times the original force.

PHYSICS!!
To expand upon bobpursley's solution: P = W/t =(1/2 m v^2)/t = (m v^2)/(2t) m = 1300 kg, v = 30 m/s, t = 6 s The work is equal to the change in kinetic energy, which, since it started from rest, is simply (1/2) m v^2 The formula given is the definition of power.

physics
In general, v = v_0 + a t, where a is any acceleration. Let t1 be the time during the acceleration and t2 be the time during the deceleration and v = the speed at the end of the acceleration. v - 0 = a t1 0 - v = -b t2 v = a t1 = b t2 This yields the equation a t1 - b t2 = 0 ...

physics
Your given answer for (2) can't be correct, it doesn't have units of distance (it has units of acceleration).

Physics
Also (and this contributes only scantly), the force of gravity is diminished as the rocket moves farther from Earth.

physics
Sorry! I just noticed your answer had not been converted to km/h. You don't show how you got your answer, so it's impossible to tell where you might have made an error. The answer I get, 29.45 m/2, is close to yours.

physics
That can't be correct. The speeder must be traveling faster than the police car (90 km/h).

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