B. It is the only graph that accurately describes acceleration as speed gradually increases.
The system of two rods will lie on the table as show in the figure
The center of first rod will lie exactly at the edge of first rod and then the center of mass of two combined rods will lie at the edge of the table.
So now the whole system will rest on the rod and it will not tipping off.
Since both rods are identical so we can say that the system will have its center of mass at the mid point on the line joining the two centers
So the value of x will be at mid point of line joining the two points on rod
![x = \frac{12.5}{2} = 6.25 cm](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B12.5%7D%7B2%7D%20%3D%206.25%20cm)
So total length over the edge will be given as
![L = 6.25 cm + 12.5 cm = 18.75 cm](https://tex.z-dn.net/?f=L%20%3D%206.25%20cm%20%2B%2012.5%20cm%20%3D%2018.75%20cm)
To solve this problem we need to use the equations related to Elastic Potential Energy, which allows us to know the Energy stored in a spring based on its elastic constant and its respective compression. By definition it is described as
![PE = \frac{1}{2}kx^2](https://tex.z-dn.net/?f=PE%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
Where,
k = Spring constant
x = Displacement
Our values are given as
As we do not know the spring constant but if the energy stored at the compression given then,
![PE = \frac{1}{2}kx^2](https://tex.z-dn.net/?f=PE%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
![20 = \frac{1}{2}k(0.08)^2](https://tex.z-dn.net/?f=20%20%3D%20%5Cfrac%7B1%7D%7B2%7Dk%280.08%29%5E2)
![k = 6250N/m](https://tex.z-dn.net/?f=k%20%3D%206250N%2Fm%20)
In the case of the second compression and understanding that the spring constant is intrinsic to the internal force of the spring, then
![x_2 = 0.12m](https://tex.z-dn.net/?f=x_2%20%3D%200.12m)
Replacing,
![PE = \frac{1}{2}(6250)(0.12)^2](https://tex.z-dn.net/?f=PE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%286250%29%280.12%29%5E2)
![PE = 45J](https://tex.z-dn.net/?f=PE%20%3D%2045J)
Therefore the elastic potential energy in the spring at this elongation is 45J
Answer:
I think is D
Explanation:
Please mark me as brainliest
Answer:
e. The speed of the object is a constant 6 m/s
Explanation:
Since the net force is towards the centre , hence there is no tangential acceleration . Only centripetal acceleration is there . Hence point mass is moving with uniform speed . Let it be u .
Centripetal force = m v² / r , r is radius of circular path .
Putting the given values
.10 x v² / .36 = 10
v = 6 m /s