To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).
The answer is 36 kg m/s
Answer:
Explanation:
The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.
Change in gravity potential energy = change in spring potential energy
mgh = 1/2kh^2
Assume gravity constant g is 10m/s^2
2.6*10*h = 1/2*200*h^2
100h^2 - 26h = 0
2h(50h - 13) = 0
h = 0 or h = 13/50 = 0.65m
h = 0 is before the spring is stretched
So the maximum distance is 0.65m.
It is helpful to study the light that comes off stars because C. The light from a star gives hints of what elements make up a star.
Answer:
the pressure at B is 527psf
Explanation:
Angular velocity, ω = v / r
ω = 20 /1.5
= 13.333 rad/s
Flow equation from point A to B
![P_A+rz_A-\frac{1}{2} Pr_A^2w^2=P_B+rz_B-\frac{1}{2} pr^2_Bw^2\\\\P_B = P_A + r(z_A-z_B)+\frac{1}{2} pw^2[(r_B^2)-(r_A)^2]\\\\P_B = [25 +(0.8+62.4)(0-1)+\frac{1}{2}(0.8\times1.94)\times(13.333)^2[2.5^2-1.5^2] ]\\\\P_B = 25 - 49.92+551.79\\\\P_B = 526.87psf\\\approx527psf](https://tex.z-dn.net/?f=P_A%2Brz_A-%5Cfrac%7B1%7D%7B2%7D%20Pr_A%5E2w%5E2%3DP_B%2Brz_B-%5Cfrac%7B1%7D%7B2%7D%20pr%5E2_Bw%5E2%5C%5C%5C%5CP_B%20%3D%20P_A%20%2B%20r%28z_A-z_B%29%2B%5Cfrac%7B1%7D%7B2%7D%20pw%5E2%5B%28r_B%5E2%29-%28r_A%29%5E2%5D%5C%5C%5C%5CP_B%20%3D%20%5B25%20%2B%280.8%2B62.4%29%280-1%29%2B%5Cfrac%7B1%7D%7B2%7D%280.8%5Ctimes1.94%29%5Ctimes%2813.333%29%5E2%5B2.5%5E2-1.5%5E2%5D%20%20%5D%5C%5C%5C%5CP_B%20%3D%2025%20-%2049.92%2B551.79%5C%5C%5C%5CP_B%20%3D%20526.87psf%5C%5C%5Capprox527psf)
the pressure at B is 527psf
