Recent content by geosonel

yes (at the location of the pendulum)

you've already calculated the gravity "g" for both planets. (note: to obtain g in mks units, recalulate values using pendulum length of 0.5 meters instead of 50 cm.)

remember that linear velocity (v), linear acceleration (a), angular velocity (ω), and angular acceleration (α) are all vector quantities. when solving a problem, one of the first jobs is to select convenient (orthogonal) coordinate axes into which these vector quantities can be projected into...

for single-slit diffraction, the condition for MINIMUMs is: y \ = \ \frac{m \lambda D}{s} where y is the distance from central max to the m-th minimum of the diffraction pattern, m a positive integer, λ the wavelength, D the distance between single-slit and distant observation screen...

since the drum rolls without slipping: Angular Acceleration = α = a/R to understand this, what's the relationship between the linear motion of the drum's axis (moving down the ramp) to the rolling outer surface circumference? remember, there's no slipping.

check here: https://www.physicsforums.com/showthread.php?t=83812

from the stated problem, it seems you'd perceive an apparent change in frequency, not in wavelength. it seems like you'd be able to see the physical extent of the wave and be able to measure the wavelength directly during each strobe pulse. thus, you'd not detect any apparent change in...

an AC electric generator uses electromagnetic induction in a coil (rotating in a magnetic field) which is described by Faraday's Law. briefly, the changing magnetic flux thru a coil being rotated in a magnetic filed induces an AC voltage in the coil. the magnetic flux thru the coil is...

unless there's more to this problem than you indicated, it appears you're correct that the answer SHOULD be (W2/W1)=2.

there are 2 angles involved in this problem. first, the 2 polarizers are rotated relative to one another by angle θ2=38.8° then, the plane polarized incident light is rotated relative to the first polarizer by angle θ1. the total net transmission thru both polarizers is given to be 0.146, so...

if you need review of solving inequality equations, try here: http://www.sosmath.com/algebra/inequalities/ineq01/ineq01.html proceed page by page thru the various review topics to the extent needed.

you can. however, what are the solutions to the 2 inequalities below (for p > 0)? (1-200/p) < 0 (1-200/p) > 0 and under what condition does the following inequality hold: (1-p/1000) > 0 the above inequality solutions give the 2 ranges of p.

you're thinking it's more difficult than it really is. begin by evaluating (dp/dt) at p=100 and p=500. which term changes sign? do any other terms change sign? now generalize these observations over the 2 given ranges of p.

begin with your correct equation: 2\frac{dv}{dt} \ = \ 16x - 16-v^2 rearrange terms: \frac{dv}{dt} + \frac{v^2}{2} \ = \ 8x - 8 then use: \frac{dv}{dt} \ = \ \frac{dv}{dx} \cdot \frac{dx}{dt} \ = \ \frac{dv}{dx} \cdot v so that placing this result into the diff eq...

1) set d2P/dt2 = 0 and solve for t is a good beginning. 2) show that (dp/dt)>0 when 200<P<1000, and (dp/dt)<0 when 0<P<200. it's not difficult.