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CANTIUS
Answers (21)
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.
That can't be correct. The speeder must be traveling faster than the police car (90 km/h).
Also (and this contributes only scantly), the force of gravity is diminished as the rocket moves farther from Earth.
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
Your given answer for (2) can't be correct, it doesn't have units of distance (it has units of acceleration).
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
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.
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
m = 29 kg v = 7.9 m/s
You solved it! However, the problem is not asking for the mass, but rather the weight w = 6664.6 N
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.
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
P = F v = w v = (800+600)(4) = 5600 W
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]
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.
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
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
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
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.
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
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
