Answer:
-0.4454 Joules
Explanation:
m = Mass of block = 2 kg
h = Height of extension = 17 cm = x
g = Acceleration due to gravity = 9.81 m/s²
Potential energy of the spring
![P=mgh\\\Rightarrow P=2\times 9.81\times 0.17\\\Rightarrow P=3.3354\ J](https://tex.z-dn.net/?f=P%3Dmgh%5C%5C%5CRightarrow%20P%3D2%5Ctimes%209.81%5Ctimes%200.17%5C%5C%5CRightarrow%20P%3D3.3354%5C%20J)
The kinetic energy of the spring
![K=\frac{1}{2}mx^2\\\Rightarrow K=\frac{1}{2}\times 200\times 0.17^2\\\Rightarrow K=2.89\ J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmx%5E2%5C%5C%5CRightarrow%20K%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20200%5Ctimes%200.17%5E2%5C%5C%5CRightarrow%20K%3D2.89%5C%20J)
In this system as the potential and kinetic energy is conserved from work energy equivalence we get
![W=P-K\\\Rightarrow W=2.89-3.3354\\\Rightarrow W=-0.4454\ J](https://tex.z-dn.net/?f=W%3DP-K%5C%5C%5CRightarrow%20W%3D2.89-3.3354%5C%5C%5CRightarrow%20W%3D-0.4454%5C%20J)
The work done by friction is -0.4454 Joules
The acceleration is the principal subordinate of the speed if the speed is steady the subsidiary is invalid if the speed is diminishing the subsidiary is negative. When discussing so much stuff we consider the momentary esteem.
<span>Note that when you back off, you back off by and large yet can locally in time quicken a tiny bit, suppose amid 1/tenth of a sec since you achieved a segment of the street which was slanting. In any case, this does not change the way that when the speed diminishes, the quickening is negative.</span>
Answer:
11, 760 Pa.
Explanation:
By applying formula P= pgh, where P is pressure, p is density, g is gravitational acceleration (9.8 m\s2) and h is height of water level. Putting values in the formula, you can have the correct answer.
Solution :
We know that :
Formula for Gravitational force is given by :
![$F_g=\frac{Gmn}{r^2}$](https://tex.z-dn.net/?f=%24F_g%3D%5Cfrac%7BGmn%7D%7Br%5E2%7D%24)
where, G is the gravitational constant
M is the mass of the bigger body
m is the mass of the smaller body
r is the distance between the two bodies.
And the formula for the centripetal force is given by :
![$F_c=\frac{mv^2}{r}$](https://tex.z-dn.net/?f=%24F_c%3D%5Cfrac%7Bmv%5E2%7D%7Br%7D%24)
where, m is the mass of the rotating body
v is the velocity
r is the radius of rotation of the body.
We know that mathematically, the gravitational force is equal to the centripetal force of the body.
Therefore,
![$F_g=F_c$](https://tex.z-dn.net/?f=%24F_g%3DF_c%24)
![$\frac{GMm}{r^2}=\frac{mv^2}{r}$](https://tex.z-dn.net/?f=%24%5Cfrac%7BGMm%7D%7Br%5E2%7D%3D%5Cfrac%7Bmv%5E2%7D%7Br%7D%24)
![$\sqrt{\frac{GM}{r}}=v$](https://tex.z-dn.net/?f=%24%5Csqrt%7B%5Cfrac%7BGM%7D%7Br%7D%7D%3Dv%24)
Hence derived.