Recent content by lemin_rew

Thank you so much for your time and help. i really appreciate it!:)

yes r1 and r2 is are the distances from point p. and the angle is between the midpoint of d(distance of the slits) to p. it was needed for the other question for calculating dsinθ. i don't think the angle is involved in the question i posted here.

so then, when m=0 gives the biggest wavelength. hence, 0.8cm=(0+1/2)λ λ=1.6cm. is this correct?

1? so then m=1 and the equation becomes Δr=(m+1/2)λ 0.8cm (1+1/2)λ λ=0.53?? is this right??...

so...then the waves are now in phase? sorry i don't think I am understanding it...fully

it is destructive because the it has nodes(minimum amplitude), and they are out of phase. the path difference Δr need to be the multiple of λ? I am not sure how to approach from now on...

Homework Statement Assume that the waves from the sources are emitted in phase, and that at the point P, amplitude of the disturbance is zero. what is the wavelength of the traveling wave emitted by the sources? (there is more than one possible answer here,but provide the largest possible...

i got my parabola to be 22.18v1f^2-0.257v1f-26.6463 and my root is 1.10. its not the correct answer...

oh i think i get what you mean, let me try thank you

so, then kinetic energy plays a role and the equation is: 1/2m1v1i^2 + 1/2m2v2i = 1/2m1v1f^2 + 1/2m2v2f^2 where v1i and v2i have opposite signs?

would it be then: V1f=((m1-m2)/(m1+m2))v1i + ((2m2)/(m1+m2))/v2i

oh okay! thank you so much. yes i got the answer. then would the second part be the same equation but with the opposite direction. so, m1*V1i + m2*-V2i = m1*V1f + m2*-V2f like this?

Homework Statement a)Two blocks are moving on a frictionless surface. The first block's mass is 1.07kg and its initial velocity is 5.98m/s. The second block's mass is 2.26kg and its initial velocity is 2.59m/s. What is the velocity of the 1.07kg block after the collision if the velocity of...

Homework Statement part a and b are solved(is it correct), but I am not sure how to solve part c. A single constant force F = (3.02i + 5.45j)N acts on a 3.69kg particle. a) Calculate the work done by this force if the particle moves from the origin to the point having the vector...

using this equation, gives me 0.0729, whereas the previous one i used gives me 0.1032. so, i can conclude that Δw = √((yΔx)² + (xΔy)²) formula gives me more precise uncertainty?