Answer:
9.9 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
![v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 5+0^2}\\\Rightarrow v=9.9\ m/s](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B2as%2Bu%5E2%7D%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B2%5Ctimes%209.81%5Ctimes%205%2B0%5E2%7D%5C%5C%5CRightarrow%20v%3D9.9%5C%20m%2Fs)
If the body has started from rest then the initial velocity is 0. In order to find the velocity just before hitting the water then the distance at which the downward motion stops is irrelevant.
Hence, the speed of the diver just before striking the water is 9.9 m/s
Answer:
A
Explanation: A is a example of Air resistance force, which is a contact force.
Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
Determine Fx."
![F_{x}=-1N.m](https://tex.z-dn.net/?f=F_%7Bx%7D%3D-1N.m)
Explanation:
We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.
torque=cross product of force and position . mathematically this can be express as
![T=r*F](https://tex.z-dn.net/?f=T%3Dr%2AF)
Where
and the position vector
![r=(2m)i-(3m)j+(2m)k](https://tex.z-dn.net/?f=r%3D%282m%29i-%283m%29j%2B%282m%29k)
using the determinant method to expand the cross product in order to determine the torque we have
![\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2%26-3%262%5C%5C%20F_%7Bx%7D%20%267%26-5%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C)
by expanding we arrive at
![T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\](https://tex.z-dn.net/?f=T%3D%2818-14%29i-%28-12-2F_%7Bx%7D%29j%2B%2812%2B3F_%7Bx%7D%29k%5C%5CT%3D4i-%28-12-2F_%7Bx%7D%29j%2B%2812%2B3F_%7Bx%7D%29k%5C%5C%5C%5C)
since we have determine the vector value of the toque, we now compare with the torque value given in the question
![(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\](https://tex.z-dn.net/?f=%284Nm%29i%2B%2810Nm%29j%2B%2811Nm%29k%3D4i-%28-12-2F_%7Bx%7D%29j%2B%2812%2B3F_%7Bx%7D%29k%5C%5C)
if we directly compare the j coordinate we have
![10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m](https://tex.z-dn.net/?f=10%3D-%28-12-2F_%7Bx%7D%29%5C%5C10%3D12%2B2F_%7Bx%7D%5C%5C%2010-12%3D2F_%7Bx%7D%5C%5C%20F_%7Bx%7D%3D-1N.m)
I think it 32, but i’m not sure