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)
Elements<span> in the same </span>group<span> in the periodic table </span>have similar chemical properties<span>. This is because their atoms </span>have<span> the same number of electrons in the highest occupied energy level. </span>Group<span> 1 </span>elements<span> are reactive metals called the alkali metals.</span>Group<span> 0 </span>elements<span> are unreactive non-metals called the noble gases.
</span>
Heat
gained in a system can be calculated by multiplying the given mass to the
specific heat capacity of the substance and the temperature difference. It is
expressed as follows:<span>
Heat = mC(T2-T1)
345.2 = 89.5(C)(305 - 285)
C = 0.1928 </span>J/g•K
Answer:
C.) 1.5 kg
Explanation:
Start with the equation:
![F_n_e_t=ma](https://tex.z-dn.net/?f=F_n_e_t%3Dma)
Plug in what you know, and solve:
![18=m(12)\\m=1.5kg](https://tex.z-dn.net/?f=18%3Dm%2812%29%5C%5Cm%3D1.5kg)
Find matching soluation:
C.) 1.5 kg
The answer is well log data, it is a detailed log of information taken from a borehole which geologist used to study geological formations of the earth's layer taken from samples returned from the borehole which was dugged.