Answer:
you don't have to its your choice whether you want to or not.
Explanation:
but you can not leave the fact that there dieing
Answer:
![\Delta K = 2164.053\,J](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%202164.053%5C%2CJ)
Explanation:
Let consider the observer as an inertial reference frame. The object is modelled after the Principle of Momentum Conservation:
![(32\,kg)\cdot (26\,\frac{m}{s} ) = (5.333\,kg)\cdot (0\,\frac{m}{s} )+(26.665\,kg )\cdot v](https://tex.z-dn.net/?f=%2832%5C%2Ckg%29%5Ccdot%20%2826%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%20%3D%20%285.333%5C%2Ckg%29%5Ccdot%20%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%2B%2826.665%5C%2Ckg%20%29%5Ccdot%20v)
The speed of the more massive piece is:
![v = 31.202\,\frac{m}{s}](https://tex.z-dn.net/?f=v%20%3D%2031.202%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
The kinetic energy added to the system is:
![\Delta K = \frac{1}{2}\cdot [(5.333\,kg)\cdot (0\,\frac{m}{s} )^{2}+(26.665\,kg )\cdot (31.202\,\frac{m}{s} )^{2}]-\frac{1}{2}\cdot (32\,kg)\cdot (26\,\frac{m}{s} )^{2}](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5B%285.333%5C%2Ckg%29%5Ccdot%20%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%2B%2826.665%5C%2Ckg%20%29%5Ccdot%20%2831.202%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%5D-%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%2832%5C%2Ckg%29%5Ccdot%20%2826%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D)
![\Delta K = 2164.053\,J](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%202164.053%5C%2CJ)
Answer:
M = 1433.5 kg
Explanation:
This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,
B = ρ g V
with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition
Σ F = 0
B-W = 0
B = W
body weight
W = M g
the volume is
V = l to h
rho_liquid g (l to h) = M g
M = rho_liquid l a h
we calculate
M = 1000 4.7 6.10 0.05
M = 1433.5 kg
S.I. Unit of mass is Kilogram which is denoted by Kg
In short, Your Answer would be Option C
Hope this helps!
Here in crash test the two forces are acting on the dummy in two different directions
As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.
So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors
so we can say
![F_{net} = \sqrt{F_1^2 + F_2^2}](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20%5Csqrt%7BF_1%5E2%20%2B%20F_2%5E2%7D)
here given that
![F_1 = 130.0 N](https://tex.z-dn.net/?f=F_1%20%3D%20130.0%20N)
![F_2 = 4500.0 N](https://tex.z-dn.net/?f=F_2%20%3D%204500.0%20N)
now we will plug in all data in the above equation
![F_{net} = \sqrt{4500^2 + 130^}](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20%5Csqrt%7B4500%5E2%20%2B%20130%5E%7D)
![F_{net} = 4501.9 N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%204501.9%20N)
so it will have net force 4501.9 N which will be reported by sensor