Answer:
The force constant is ![k =1.316 *10^{7} \ N/m](https://tex.z-dn.net/?f=k%20%3D1.316%20%2A10%5E%7B7%7D%20%5C%20%20N%2Fm)
The energy stored in the spring is ![E = 1.68 *10^{7} \ J](https://tex.z-dn.net/?f=E%20%3D%20%201.68%20%2A10%5E%7B7%7D%20%5C%20J)
Explanation:
From the question we are told that
The mass of the object is ![M = 4.8*10^{5} \ kg](https://tex.z-dn.net/?f=M%20%20%3D%204.8%2A10%5E%7B5%7D%20%5C%20kg)
The period is ![T = 1.2 \ s](https://tex.z-dn.net/?f=T%20%20%3D%201.2%20%5C%20s)
The period of the spring oscillation is mathematically represented as
![T =2 \pi \sqrt{ \frac{M}{k}}](https://tex.z-dn.net/?f=T%20%20%3D2%20%5Cpi%20%5Csqrt%7B%20%5Cfrac%7BM%7D%7Bk%7D%7D)
where k is the force constant
So making k the subject
![k = \frac{4 \pi ^2 M }{T^2}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B4%20%5Cpi%20%5E2%20M%20%7D%7BT%5E2%7D)
substituting values
![k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B4%20%283.142%29%20%5E2%20%284.8%20%2A10%5E%7B5%7D%29%20%7D%7B%281.2%29%5E2%7D)
![k =1.316 *10^{7} \ N/m](https://tex.z-dn.net/?f=k%20%3D1.316%20%2A10%5E%7B7%7D%20%5C%20%20N%2Fm)
The energy stored in the spring is mathematically represented as
![E = \frac{1}{2} k x^2](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20k%20x%5E2)
Where x is the spring displacement which is given as
![x = 1.6 \ m](https://tex.z-dn.net/?f=x%20%3D%20%201.6%20%5C%20m)
substituting values
![E = \frac{1}{2} (1.316 *10^{7}) (1.6)^2](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%281.316%20%2A10%5E%7B7%7D%29%20%281.6%29%5E2)
![E = 1.68 *10^{7} \ J](https://tex.z-dn.net/?f=E%20%3D%20%201.68%20%2A10%5E%7B7%7D%20%5C%20J)
Answer:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Explanation:
The bulk modulus is represented by the following differential equation:
![K = - V\cdot \frac{dP}{dV}](https://tex.z-dn.net/?f=K%20%3D%20-%20V%5Ccdot%20%5Cfrac%7BdP%7D%7BdV%7D)
Where:
- Bulk module, measured in pascals.
- Sample volume, measured in cubic meters.
- Local pressure, measured in pascals.
Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:
![-\frac{K \,dV}{V} = dP](https://tex.z-dn.net/?f=-%5Cfrac%7BK%20%5C%2CdV%7D%7BV%7D%20%3D%20dP)
This resultant expression is solved by definite integration and algebraic handling:
![-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP](https://tex.z-dn.net/?f=-K%5Cint%5Climits%5E%7BV_%7Bf%7D%7D_%7BV_%7Bo%7D%7D%20%7B%5Cfrac%7BdV%7D%7BV%7D%20%7D%20%3D%20%5Cint%5Climits%5E%7BP_%7Bf%7D%7D_%7BP_%7Bo%7D%7D%5C%2C%20dP)
![-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}](https://tex.z-dn.net/?f=-K%5Ccdot%20%5Cln%20%5Cleft%20%7C%5Cfrac%7BV_%7Bf%7D%7D%7BV_%7Bo%7D%7D%20%5Cright%7C%20%3D%20P_%7Bf%7D%20-%20P_%7Bo%7D)
![\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%7C%20%5Cfrac%7BV_%7Bf%7D%7D%7BV_%7Bo%7D%7D%20%5Cright%7C%20%3D%20%5Cfrac%7BP_%7Bo%7D-P_%7Bf%7D%7D%7BK%7D)
![\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7Bf%7D%7D%7BV_%7Bo%7D%7D%20%3D%20e%5E%7B%5Cfrac%7BP_%7Bo%7D-P_%7Bf%7D%7D%7BK%7D%20%7D)
The final volume is predicted by:
![V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }](https://tex.z-dn.net/?f=V_%7Bf%7D%20%3D%20V_%7Bo%7D%5Ccdot%20e%5E%7B%5Cfrac%7BP_%7Bo%7D-P_%7Bf%7D%7D%7BK%7D%20%7D)
If
,
and
, then:
![V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }](https://tex.z-dn.net/?f=V_%7Bf%7D%20%3D%20%281%5C%2Cm%5E%7B3%7D%29%20%5Ccdot%20e%5E%7B%5Cfrac%7B-10.1325%5Ctimes%2010%5E%7B6%7D%5C%2CPa%7D%7B2.3%20%5Ctimes%2010%5E%7B9%7D%5C%2CPa%7D%20%7D)
![V_{f} \approx 0.996\,m^{3}](https://tex.z-dn.net/?f=V_%7Bf%7D%20%5Capprox%200.996%5C%2Cm%5E%7B3%7D)
Change in volume due to increasure on pressure is:
![\Delta V = V_{o} - V_{f}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20V_%7Bo%7D%20-%20V_%7Bf%7D)
![\Delta V = 1\,m^{3} - 0.996\,m^{3}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%201%5C%2Cm%5E%7B3%7D%20-%200.996%5C%2Cm%5E%7B3%7D)
![\Delta V = 0.004\,m^{3}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%200.004%5C%2Cm%5E%7B3%7D)
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Answer:33
Explanation:
F = frequency
N = Node count
w = wave lenght
v = wave velocity
L = distance wave traveled
First find wave length of laser
w = (2/(N))*(L)
w = (2/(10))*(8)
w = 1.6
then using (w), find velocity
V = (w)(F)
V = (1.6)*(108)
V = 288
Plug in V and the new frequency to solve for new node count
F = NV/2L
(600) = (N)*(288) / 2 * (8)
(N) = 33.33
there are 33 nodes
Answer:
A blackbody, or Planckian radiator, is a cavity within a heated material from which heat cannot escape. No matter what the material, the walls of the cavity exhibit a characteristic spectral emission, which is a function of its temperature.
Example:
Emission from a blackbody is temperature dependent and at high temperature, a blackbody will emit a spectrum of photon energies that span the visible range, and therefore it will appear white. The Sun is an example of a high-temperature blackbody.