Answer:
The force must he apply to the sled is of F= 764.4 N.
Explanation:
m= 60 kg
g= 9.8 m/s²
μ=0.3
W= m*g
W= 588 N
Fr= μ*W
Fr= 176.4 N
F= W + Fr
F= 764.4 N
The solution for this is:
Work done = force * distance = m*a*d and power = energy/time
The vo=0 and vf = 25 m/s and t=7 sec. This gives...
3.6 m/s^2 as acceleration and d=87.5 meters and thus F=ma= 5400 N.
Energy = 5400*87.5 = 4.7E5 Joules (2 sig. figs) and Power = 67,500 Watts or 90 HP (2 sig. figs again).
In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
![y= \frac{m \lambda D}{d}](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7Bm%20%5Clambda%20D%7D%7Bd%7D%20)
where D=5.00 m is the distance of the screen from the slits, and
![d=0.048 mm=0.048 \cdot 10^{-3}m](https://tex.z-dn.net/?f=d%3D0.048%20mm%3D0.048%20%5Ccdot%2010%5E%7B-3%7Dm)
is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
![\lambda = \frac{yd}{mD}= \frac{(0.065 m)(0.048 \cdot 10^{-3}m)}{(1)(5.00 m)}= 6.24 \cdot 10^{-7}m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%20%5Cfrac%7Byd%7D%7BmD%7D%3D%20%5Cfrac%7B%280.065%20m%29%280.048%20%5Ccdot%2010%5E%7B-3%7Dm%29%7D%7B%281%29%285.00%20m%29%7D%3D%20%206.24%20%5Ccdot%2010%5E%7B-7%7Dm)
And from the relationship between frequency and wavelength,
![c=\lambda f](https://tex.z-dn.net/?f=c%3D%5Clambda%20f)
, we can find the frequency of the light:
Central maximum = d* wavelength/ D
thus
12*10-^3 = 3.4*6.32*10-^7/D
D = 3.4*6.32*10-^7/12*10-^3
D = 1.79*10-^4 m