Answer:
e) 11 m/s
Explanation:
For a particle under the action of a conservative force, its mechanical energy at point 0 is must be equal to its mechanical energy at point 1:
![K_1+U_1=K_0+U_0\\\\\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(x_1)^2+(2.0\frac{J}{m^4})(x_1)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})(x_0)^2+(2.0\frac{J}{m^4})(x_0)^4]](https://tex.z-dn.net/?f=K_1%2BU_1%3DK_0%2BU_0%5C%5C%5C%5C%5Cfrac%7Bmv_1%5E2%7D%7B2%7D%2B%5B%288.0%5Cfrac%7BJ%7D%7Bm%5E2%7D%29%28x_1%29%5E2%2B%282.0%5Cfrac%7BJ%7D%7Bm%5E4%7D%29%28x_1%29%5E4%5D%3D%5Cfrac%7Bmv_0%5E2%7D%7B2%7D%2B%5B%288.0%5Cfrac%7BJ%7D%7Bm%5E2%7D%29%28x_0%29%5E2%2B%282.0%5Cfrac%7BJ%7D%7Bm%5E4%7D%29%28x_0%29%5E4%5D)
In 
the speed is given, so 
and 
. Replacing:
![\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(1m)^2+(2.0\frac{J}{m^4})(1m)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})0^2+(2.0\frac{J}{m^4})(0)^4]\\\frac{mv_1^2}{2}+8.0J+2.0J=\frac{mv_0^2}{2}\\\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+10J\\v_0=\sqrt{\frac{2}{m}(\frac{mv_1^2}{2}+10J)}\\v_0=\sqrt{\frac{2}{0.2kg}(\frac{(0.2kg)(5\frac{m}{s})^2}{2}+10J)}\\v_0=11.18\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv_1%5E2%7D%7B2%7D%2B%5B%288.0%5Cfrac%7BJ%7D%7Bm%5E2%7D%29%281m%29%5E2%2B%282.0%5Cfrac%7BJ%7D%7Bm%5E4%7D%29%281m%29%5E4%5D%3D%5Cfrac%7Bmv_0%5E2%7D%7B2%7D%2B%5B%288.0%5Cfrac%7BJ%7D%7Bm%5E2%7D%290%5E2%2B%282.0%5Cfrac%7BJ%7D%7Bm%5E4%7D%29%280%29%5E4%5D%5C%5C%5Cfrac%7Bmv_1%5E2%7D%7B2%7D%2B8.0J%2B2.0J%3D%5Cfrac%7Bmv_0%5E2%7D%7B2%7D%5C%5C%5Cfrac%7Bmv_0%5E2%7D%7B2%7D%3D%5Cfrac%7Bmv_1%5E2%7D%7B2%7D%2B10J%5C%5Cv_0%3D%5Csqrt%7B%5Cfrac%7B2%7D%7Bm%7D%28%5Cfrac%7Bmv_1%5E2%7D%7B2%7D%2B10J%29%7D%5C%5Cv_0%3D%5Csqrt%7B%5Cfrac%7B2%7D%7B0.2kg%7D%28%5Cfrac%7B%280.2kg%29%285%5Cfrac%7Bm%7D%7Bs%7D%29%5E2%7D%7B2%7D%2B10J%29%7D%5C%5Cv_0%3D11.18%5Cfrac%7Bm%7D%7Bs%7D)
The solution would be like this for this specific problem:
Given:
diffraction grating
slits = 900 slits per centimeter
interference pattern that
is observed on a screen from the grating = 2.38m
maxima for two different
wavelengths = 3.40mm
slit separation .. d =
1/900cm = 1.11^-3cm = 1.111^-5 m <span>
Whenas n = 1, maxima (grating equation) sinθ = λ/d
Grant distance of each maxima from centre = y ..
<span>As sinθ ≈ y/D y/D =
λ/d λ = yd / D </span>
∆λ = (λ2 - λ1) = y2.d/D - y1.d/D
∆λ = (d/D) [y2 -y1]
<span>∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m</span></span>
Answer:
Speed is 0.08 m/s.
Explanation:
Given the distance that the bird flies = 3.7 meters
The time is taken by the bird to fly the 3.7 meters = 46 seconds
We have given distance and time. Now we have to find the speed at which the bird flies. So, to calculate the speed of the bird we have to divide the distance by the time.
Below is the formula to find the speed.
Speed = Distance / Time
Now insert the given value in the formula.
Speed = 3.7 / 46 = 0.08 m/s
