Answer:
![v\approx 8.570\,\frac{m}{s}](https://tex.z-dn.net/?f=v%5Capprox%208.570%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
Explanation:
The equation of equlibrium for the box is:
![\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a](https://tex.z-dn.net/?f=%5CSigma%20F_%7Bx%7D%20%3D%2018%5C%2CN-%280.530%5C%2C%5Cfrac%7BN%7D%7Bm%7D%20%29%5Ccdot%20x%20%3D%20%287.90%5C%2Ckg%29%5Ccdot%20a)
The formula for the acceleration, given in
, is:
![a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B18%5C%2CN-%280.530%5C%2C%5Cfrac%7BN%7D%7Bm%7D%20%29%5Ccdot%20x%7D%7B7.90%5C%2Ckg%7D)
Velocity can be derived from the following definition of acceleration:
![a = v\cdot \frac{dv}{dx}](https://tex.z-dn.net/?f=a%20%3D%20v%5Ccdot%20%5Cfrac%7Bdv%7D%7Bdx%7D)
![v\, dv = a\, dx](https://tex.z-dn.net/?f=v%5C%2C%20dv%20%3D%20a%5C%2C%20dx)
![\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20v%5E%7B2%7D%20%3D%20%5Cint%5Climits%5E%7B17%5C%2Cm%7D_%7B0%5C%2Cm%7D%20%7B%5Cfrac%7B18%5C%2CN-%280.530%5C%2C%5Cfrac%7BN%7D%7Bm%7D%20%29%5Ccdot%20x%7D%7B7.90%5C%2Ckg%7D%20%7D%20%5C%2C%20dx)
![\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20v%5E%7B2%7D%20%3D%5Cfrac%7B18%5C%2CN%7D%7B7.90%5C%2Ckg%7D%20%20%5Cint%5Climits%5E%7B17%5C%2Cm%7D_%7B0%5C%2Cm%7D%5C%2C%20dx%20%20-%20%5Cfrac%7B0.530%5C%2C%5Cfrac%7BN%7D%7Bm%7D%20%7D%7B7.90%5C%2Ckg%7D%20%5Cint%5Climits%5E%7B17%5C%2Cm%7D_%7B0%5C%2Cm%7D%20%7Bx%7D%20%5C%2C%20dx)
![\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20v%5E%7B2%7D%20%3D%20%282.278%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20x%20%7C_%7B0%5C%2Cm%7D%5E%7B27%5C%2Cm%7D-%280.034%5C%2C%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20x%5E%7B2%7D%7C_%7B0%5C%2Cm%7D%5E%7B27%5C%2Cm%7D)
![v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }](https://tex.z-dn.net/?f=v%20%3D%5Csqrt%7B2%5Ccdot%5B%282.278%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20x%20%7C_%7B0%5C%2Cm%7D%5E%7B27%5C%2Cm%7D-%280.034%5C%2C%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20x%5E%7B2%7D%7C_%7B0%5C%2Cm%7D%5E%7B27%5C%2Cm%7D%5D%20%20%7D)
The speed after the box has travelled 17 meters is:
![v\approx 8.570\,\frac{m}{s}](https://tex.z-dn.net/?f=v%5Capprox%208.570%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
D is correct, Gamma rays have the shortest wavelength.
The rain gets evaporated in to water vapor and is returned to the clouds where they go through condensation and then they poud down as rain or A.K.A, Precipitation.
Answer:
Explanation:
a ) Let let the frictional force needed be F
Work done by frictional force = kinetic energy of car
F x 107 = 1/2 x 1400 x 35²
F = 8014 N
b )
maximum possible static friction
= μ mg
where μ is coefficient of static friction
= .5 x 1400 x 9.8
= 6860 N
c )
work done by friction for μ = .4
= .4 x 1400 x 9.8 x 107
= 587216 J
Initial Kinetic energy
= .5 x 1400 x 35 x 35
= 857500 J
Kinetic energy at the at of collision
= 857500 - 587216
= 270284 J
So , if v be the velocity at the time of collision
1/2 mv² = 270284
v = 19.65 m /s
d ) centripetal force required
= mv₀² / d which will be provided by frictional force
= (1400 x 35 x 35) / 107
= 16028 N
Maximum frictional force possible
= μmg
= .5 x 1400 x 9.8
= 6860 N
So this is not possible.