Answer:
v = 0
Explanation:
This problem can be solved by taking into account:
- The equation for the calculation of the period in a spring-masss system
( 1 )
- The equation for the velocity of a simple harmonic motion
( 2 )
where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block
Hence
and by reeplacing it in ( 2 ):
In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.
This question is in complete.The question is
A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.
Answer:
distance=0.124 m
Explanation:
0.495 m/s
Explanation
the formula for the terminal velocity is given by:
![\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2mg%7D%7B%5Csigma%20AC%7D%7D%20%5C%5C%20%5Ctext%7Bwhere%7D%20%5C%5C%20%20%5Cend%7Bgathered%7D)
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient
then
Step 1
Let's find the mass
now, replace
![\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2mg%7D%7B%5Csigma%20AC%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2%280.002kg%29%289.81%5Ctext%7B%20%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%7D%7B%282%5Ccdot10%5E3%5Cfrac%7B%5Coperatorname%7Bkg%7D%7D%7Bm%5E3%7D%29%280.0001m%5E2%290.8%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.03924%5Cfrac%7B%5Coperatorname%7Bkg%7Dm%7D%7Bs%5E2%7D%7D%7B0.16%5Cfrac%7B%5Coperatorname%7Bkg%7D%7D%7Bm%5E%7B%7D%7D%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B0.2452%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%7D%20%5C%5C%20v%3D0.495%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
hence, the answer is 0.495 m/s
