Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
Answer:
The speed of the laser light in the cable, ![c_f=2.1\times 10^8\ m/s](https://tex.z-dn.net/?f=c_f%3D2.1%5Ctimes%2010%5E8%5C%20m%2Fs)
Explanation:
It is given that,
Wavelength of Argon laser, ![\lambda=5\times 10^2\ nm=5\times 10^{-7}\ m](https://tex.z-dn.net/?f=%5Clambda%3D5%5Ctimes%2010%5E2%5C%20nm%3D5%5Ctimes%2010%5E%7B-7%7D%5C%20m)
Refractive index, n = 1.46
Let
is the speed of the laser light in the cable. The speed of light in a medium is given by :
![c_f=\dfrac{c}{n}](https://tex.z-dn.net/?f=c_f%3D%5Cdfrac%7Bc%7D%7Bn%7D)
![c_f=\dfrac{3\times 10^8\ m/s}{1.46}](https://tex.z-dn.net/?f=c_f%3D%5Cdfrac%7B3%5Ctimes%2010%5E8%5C%20m%2Fs%7D%7B1.46%7D)
![c_f=2.05\times 10^8\ m/s](https://tex.z-dn.net/?f=c_f%3D2.05%5Ctimes%2010%5E8%5C%20m%2Fs)
or
![c_f=2.1\times 10^8\ m/s](https://tex.z-dn.net/?f=c_f%3D2.1%5Ctimes%2010%5E8%5C%20m%2Fs)
So, the speed of the laser light is
. Hence, this is the required solution.
Answer: Preoperational stage (second stage).
Explanation: In the second stage of Child's development i.e., from the age of 2 years to 7 years approximately, which is called as Preoperational stage, a child develops a strong sense of Egocentrism. It is a situation where a child is unable to see thing from other peoples' eye and feels that everyone's reaction to something should be same as the child's. At this stage, a child is busy and completely trapped in his own dreamy world and is not ready to understand things properly.
Hence, the stage is Preoperational stage.
Answer:
To survive, every cell must have a constant supply of vital substances such as sugar, minerals, and oxygen, and dispose of waste products, all carried back and forth by the blood cells. Without these substances, cells would die in a very short period of time.
Explanation:
This problem provides information about the pressure and temperature ideal gases are studied at. The answer to the questions are that all molecules have the same density, 2.43x10²⁵ mol/m³ and 2.43x10¹⁹ mol/cm³.
<h3>Idela gases</h3>
In science, we can start studying gases with the concept of ideal gas, as they do not collide one to another and are assumed to be perfect spheres with no relevant interactions.
In such a way, one can conclude that the <u>number density of all ideal gasses at SATP is the same</u>, as they are assumed to be perfect spheres with equal volumes per molecule.
Moreover, when calculating the number of molecules per cubic meter, one must use the ideal gas equation as:
![PV=nRT\\\\\frac{N}{V}= \frac{P*N_A}{RT}](https://tex.z-dn.net/?f=PV%3DnRT%5C%5C%5C%5C%5Cfrac%7BN%7D%7BV%7D%3D%20%5Cfrac%7BP%2AN_A%7D%7BRT%7D)
And plug in the numbers we are given:
![\frac{N}{V}= \frac{100kPa*\frac{1000Pa}{1kPa}*6.022x10^{23}molec/mol}{8.314\frac{Pa*m^3}{mol*K}*298K}=2.43x10^{25}molec/m^3](https://tex.z-dn.net/?f=%5Cfrac%7BN%7D%7BV%7D%3D%20%5Cfrac%7B100kPa%2A%5Cfrac%7B1000Pa%7D%7B1kPa%7D%2A6.022x10%5E%7B23%7Dmolec%2Fmol%7D%7B8.314%5Cfrac%7BPa%2Am%5E3%7D%7Bmol%2AK%7D%2A298K%7D%3D2.43x10%5E%7B25%7Dmolec%2Fm%5E3)
Lastly, we can calculate the molecules per cubic centimeter by performing the following conversion:
![2.43x10^{25}\frac{molec}{m^3}*(\frac{1m}{100cm} )^3\\ \\=2.43x10^{19}\frac{molec}{cm^3}](https://tex.z-dn.net/?f=2.43x10%5E%7B25%7D%5Cfrac%7Bmolec%7D%7Bm%5E3%7D%2A%28%5Cfrac%7B1m%7D%7B100cm%7D%20%29%5E3%5C%5C%20%5C%5C%3D2.43x10%5E%7B19%7D%5Cfrac%7Bmolec%7D%7Bcm%5E3%7D)
Learn more about ideal gases: brainly.com/question/26450101