Answer:
d = 421.83 m
Explanation:
It is given that,
Height, h = 396.9 m
Horizontal speed, v = 46.87 m/s
We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,
![396.9=0\times t+\dfrac{1}{2}\times 9.8 t^2\\\\t=9\ s](https://tex.z-dn.net/?f=396.9%3D0%5Ctimes%20t%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%209.8%20t%5E2%5C%5C%5C%5Ct%3D9%5C%20s)
Now d is the distance covered by the cannonball. So,
![d=vt\\\\d=46.87\times 9\\\\d=421.83\ m](https://tex.z-dn.net/?f=d%3Dvt%5C%5C%5C%5Cd%3D46.87%5Ctimes%209%5C%5C%5C%5Cd%3D421.83%5C%20m)
Hence, this is the required solution.
Explanation:
36-4/4= 9 m/squared. meter per squared.
acceleration unit is meter per second Square.equation is velocity by time.for average final(36) minus initial(4)
Answer:
No.
Explanation:
The "guide to Engineering and land surveying" for professional engineers and land surveyors by the California board reviews that an unlicensed person cannot be a sole owner of an engineering business, unless there is partnership with a licensed engineer.
Ya because they’re are both 50 liters
Answer:
The block's mass should be ![3m](https://tex.z-dn.net/?f=3m)
Explanation:
Given:
Cart with mass ![m](https://tex.z-dn.net/?f=m)
From the conservation of energy before mass is added,
![\frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mv%5E%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20kA%5E%7B2%7D)
Where
amplitude of spring mass system,
spring constant
![A = v\sqrt{\frac{m}{k} }](https://tex.z-dn.net/?f=A%20%3D%20v%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%20%7D)
Now new mass
is added to the system,
![\frac{1}{2} (m +M ) v^{2} = \frac{1}{2} k A^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%28m%20%2BM%20%29%20v%5E%7B2%7D%20%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%20k%20A%5E%7B2%7D)
![A = v \sqrt{\frac{m +M }{k} }](https://tex.z-dn.net/?f=A%20%3D%20v%20%5Csqrt%7B%5Cfrac%7Bm%20%2BM%20%7D%7Bk%7D%20%7D)
Here, given in question frequency is reduced to half so we can write,
![f' = \frac{f}{2}](https://tex.z-dn.net/?f=f%27%20%3D%20%5Cfrac%7Bf%7D%7B2%7D)
Where
frequency of system before mass is added,
frequency of system after mass is added.
![\omega ' = \frac{\omega}{2}](https://tex.z-dn.net/?f=%5Comega%20%27%20%3D%20%5Cfrac%7B%5Comega%7D%7B2%7D)
![\sqrt{\frac{k}{m +M} } = \frac{\sqrt{\frac{k}{m} } }{2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%20%2BM%7D%20%7D%20%20%3D%20%5Cfrac%7B%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%20%7D%7B2%7D)
![\frac{k}{m +M } = \frac{k}{4m}](https://tex.z-dn.net/?f=%5Cfrac%7Bk%7D%7Bm%20%2BM%20%7D%20%3D%20%5Cfrac%7Bk%7D%7B4m%7D)
![M = 3m](https://tex.z-dn.net/?f=M%20%3D%203m)
Therefore, the block's mass should be ![3m](https://tex.z-dn.net/?f=3m)