1) In the first quadrant (x > 0, y > 0), the x-component of the net electric field is:
always positive.
always negative.
sometimes positive and sometimes negative.
2) In the third quadrant (x < 0, y < 0), the x-component of the net electric field is:
always positive.
always...
Four charges q1 = q3 = -q and q2 = q4 = +q, where q = 9 µC, are fixed at the corners of a square with sides a = 1.3 m.
(a) Calculate the x- and y-components of the net electric field at the midpoint M of the bottom side of the square.
(b) Find the total force exerted on q4 by the charges...
How do I find the angle? What does theta correspond to in the first place? I understand that the left and right horizontal components of the force cancel though. I tried substituting 90 degrees since the two forces are perpendicular to each other, however this didn't give me a correct...
The transverse displacement of an harmonic wave on a stretched rope is y = 0.04 cos(2.5 t - 3.3 x), where x and y are in meters and t is in seconds. A 5 meter length of this rope has a mass of 1.5 kg.
a) What is the tension in the rope?
b) At time t = 0, consider a 1/2 wavelength long...
just a quick question. a friend and i were arguing over whether momentum is represented by the letter P or the greek letter rho. does anybody know which one is correct?
Is my method of calculating the I of the system correct? I took the sum of the boy and the merry-go-round, multiplied by the square of the radius and all that divided by 2.This should give me the the I of the ystem right?
Use the equation torque/(Iw). This will give you your angular velocity of precession. Now you need to take 2pi divided by your result to get it in terms of angular frequency. Glad I could help cuz
A boy of mass m = 50 kg running with speed v = 4 m/s jumps onto the outer edge of a merry-go-round of mass M = 150 kg and radius R = 2 m, as shown in the picture above. The merry-go-round is initially at rest, and can rotate about a frictionless pivot at its center. You may assume that the...
That's a really good question... :biggrin: unfortunately the question doesn't make it clear...can you explain to me though how I would go about trying each setup-both 2 rods connected at the center or 4 rods connected. Thank you.
If there are four rods of length L, then the distance to the...
The picture shows four rods. Yeah sorry I meant MR^2, basically I used pythagorem theorem and I found that R^2 equals 2(L/2)^2 which simplifies to (L^2)/2. I still can't solve the problem though because I don't know how to define the distance from each of the point mass to the axis of rotation.
An anemometer for measuring wind speed consists of four metal cups, each of mass m = 146 g, mounted on the ends of essentially massless rods of length L = 0.3 m. The rods are at right angles to each other and the structure rigidly rotates at f = 12 rev/s. Treat the cups as point masses.
a)...
Two identical uniform beams, each having a mass of 4 kg, are symmetrically set up against each other on a floor with which they have a coefficient of static friction µ = 0.609 .
a) What is the minimum angle the beams can make with the floor without falling?
b) For the angle calculated in...