Recent content by Sarah0001

Yes, distance between adjacent maxima is lambda/2. Also, I don't understand how the path difference leads to the solution that it is (a) Yes and for (b) No for recording minimia and maxima, the solution given itself is in terms of changing phases, so I would like to understand the solution...

a) When the loudspeaker is moving along the x direction instead of the screen, I understand that the phase of which the of the reflected wave meets the loudspeaker is changing, and so when the reflected wave interferes with waves behind the loudspeaker reaching microphase, the phase of the...

The arrow is following projectile motion to the target when released from an archer's bow. v vertical = 10ms^-1 v horizontal = 50 ms^-1 resultant v = √2600 mass of arrow = 20*10^-3 I attempted to use F avg = mΔv/Δt to calculate the average force where Δt = 5*10^-3 /...

What is the convention for determining whether the diffracted angles or incidence angles are negative or positive relative to the normal incidence of the surface?

A perpendicular dropped from DE to CE??

My question is Why is the sum to infinity used as opposed to Sum to n? and How can I deduce that the sum to infinity must be used from the question?Total Distance = h + 2*Sum of Geometric progression (to infinity) h + 2*h/3 / 1-1/3 h + 2h/3 *3/2 = h + h = 2h At first I did sum to infinity...

yes..That's exactly what I did. My question is what is happening, when the egg is 'brought to a stop in a distance of 1 mm' when it hits the ground. What is the motion like then? Does it rebound, does the ground deform very slightly?

s=1 *10^-3 v^2 = u^2 + 2as a = v^2 - u^2 / 2*s = - 195 822.05 m/s^2 u=19.79 m/s F= 20*10^-3 *195 822.05 = 4000 N (1 s.f) v=0 a=?or 1/2mv^2=Fd ---->...

Thank you, so the moon just has to cross the straight line path between the Earth and Sun during the new moon phase

I meant the Moon has to be in line with the Earth and Sun

From my understanding, for an observer to see a total solar eclipse, 1) the Moon's orbit has to be near aligned with the Earth's Orbit, 2)such that the New moon's Umbra falls on Earth, and 3) the observer must be within this cast umbra. But for a partial solar eclipse, since the Penumbra of the...

Is it that work is done to overcome the nuclear forces holding the nucleus together?

surely if say R is minmum , I therefore is maxmium so we can reason using P=I^2R that when R is minmum i.e at noon then P is max, as there is a factor of squaring.

A typical graph of LDR resistance against illumination would have a negative correlation. I understand resistance of the LDR goes down with increase in Lux, as from my understanding this frees more valence electrons onto the conduction band, and so power is max when R is min. This is at noon...

Which equation do I use and why do I use it? I think it would be P = V^2/R, but why do we exclude current the from the equation?