Answer:
Explanation:
The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

Where:

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since
, so:

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

Where in this case:

Therefore:

He worked with Euler on elasticity and the development of the Euler-Bernoulli beam equation.
Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution =
radians
So, 2.73 revolutions = 
Therefore, the angular velocity of the tack is, 
Now, radial acceleration of the tack is given as:

Plug in the given values and solve for
. This gives,
![a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]](https://tex.z-dn.net/?f=a_r%3D%2817.153%5C%20rad%2Fs%29%5E2%5Ctimes%2037.7%5C%20cm%5C%5Ca_r%3D294.225%5Ctimes%2037.7%5C%20cm%2Fs%5E2%5C%5Ca_r%3D11092.28%5C%20cm%2Fs%5E2%5C%5Ca_r%3D110.9%5C%20m%2Fs%5E2%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5B1%5C%20cm%20%3D%200.01%5C%20m%5D)
Therefore, the radial acceleration of the tack is 110.9 m/s².
Answer:
Explanation:
Givens
d = 8.5 meters
vi = 0
a = 9.81
t = ?
Formula
d = vi * t + 1/2 a t^2
Solution
8.5 = 0 + 1/2 9.81 * t^2 multiply both sides by 2
8.5 = 4.095 t^2 Divide both sides by 4.095
8.5/4.095 = t^2
1.7329 = t^2 Take the square root of both sides
t = 1.316
It takes 1.316 seconds to hit the ground.