Explanation:
13 cmHg (centimeters of mercury) is the pressure at the bottom of a column of mercury 13 cm deep. It is the equivalent of about 17.3 kPa or 2.5 psi.
The rock strike the water with the speed of 15.78 m/sec.
The speed by which rock hit the water is calculated by the formula
v=![\sqrt{2gh}](https://tex.z-dn.net/?f=%20%5Csqrt%7B2gh%7D%20%20)
v=![\sqrt{2*9.8*12.7}](https://tex.z-dn.net/?f=%20%5Csqrt%7B2%2A9.8%2A12.7%7D%20%20)
v=15.78 m/sec
Hence, the rock strike the water with the speed of 15.78 m/sec.
Answer:
An object can have a displacement in the absence of any external force acting on it
Explanation:
When a object moves with a constant velocity (v), then it gets displaced in the direction of motion but the net external force experienced by the object is zero.
F external =ma
If object moves with constant velocity, acceleration is zero.
Since, a=0 ⟹F external =0
Using s=ut+ 1/2 at ^2
⟹ Displacement s=ut (∵a=0)
Hence, an object can have a displacement in the absence of any external force acting on it
Hope this helped you:)
To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,
![m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2](https://tex.z-dn.net/?f=m%3D%2089%20kg%5C%5Cx%20%3D%203.1%20m%5C%5Ct%20%3D%200.5s%5C%5Ca%20%3D%20g%20%3D%209.8m%2Fs%5E2)
Through the aforementioned formula we will have to
![v_f^2-v_i^2 = 2ax](https://tex.z-dn.net/?f=v_f%5E2-v_i%5E2%20%3D%202ax)
The particulate part of the rest, so the final speed would be
![v_f^2 = 2gx](https://tex.z-dn.net/?f=v_f%5E2%20%3D%202gx)
![v_f=\sqrt{2(9.8)(3.1)}](https://tex.z-dn.net/?f=v_f%3D%5Csqrt%7B2%289.8%29%283.1%29%7D)
![v_f = 7.79m/s](https://tex.z-dn.net/?f=v_f%20%3D%207.79m%2Fs)
Now from Newton's second law we know that
![F = ma](https://tex.z-dn.net/?f=F%20%3D%20ma)
Here,
m = mass
a = acceleration, which can also be written as a function of velocity and time, then
![F = m\frac{dv}{dt}](https://tex.z-dn.net/?f=F%20%3D%20m%5Cfrac%7Bdv%7D%7Bdt%7D)
Replacing we have that,
![F = (89)\frac{7.79}{0.5}](https://tex.z-dn.net/?f=F%20%3D%20%2889%29%5Cfrac%7B7.79%7D%7B0.5%7D)
![F = 1386.62N](https://tex.z-dn.net/?f=F%20%3D%201386.62N)
Therefore the force that the water exert on the man is 1386.62