Answer
given,
Length of the string, L = 2 m
speed of the wave , v = 50 m/s
string is stretched between two string
For the waves the nodes must be between the strings
the wavelength is given by
![\lambda = \dfrac{2L}{n}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cdfrac%7B2L%7D%7Bn%7D)
where n is the number of antinodes; n = 1,2,3,...
the frequency expression is given by
![f = n\dfrac{v}{2L}](https://tex.z-dn.net/?f=f%20%3D%20n%5Cdfrac%7Bv%7D%7B2L%7D)
now, wavelength calculation
n = 1
![\lambda_1 = \dfrac{2\times 2}{1}](https://tex.z-dn.net/?f=%5Clambda_1%20%3D%20%5Cdfrac%7B2%5Ctimes%202%7D%7B1%7D)
λ₁ = 4 m
n = 2
![\lambda_2 = \dfrac{2\times 2}{2}](https://tex.z-dn.net/?f=%5Clambda_2%20%3D%20%5Cdfrac%7B2%5Ctimes%202%7D%7B2%7D)
λ₂ = 2 m
n =3
![\lambda_3 = \dfrac{2\times 2}{3}](https://tex.z-dn.net/?f=%5Clambda_3%20%3D%20%5Cdfrac%7B2%5Ctimes%202%7D%7B3%7D)
λ₃ = 1.333 m
now, frequency calculation
n = 1
![f = n\dfrac{v}{2L}](https://tex.z-dn.net/?f=f%20%3D%20n%5Cdfrac%7Bv%7D%7B2L%7D)
![f_1 =1\times \dfrac{50}{2\times 2}](https://tex.z-dn.net/?f=f_1%20%3D1%5Ctimes%20%5Cdfrac%7B50%7D%7B2%5Ctimes%202%7D)
f₁ = 12.5 Hz
n = 2
![f = n\dfrac{v}{2L}](https://tex.z-dn.net/?f=f%20%3D%20n%5Cdfrac%7Bv%7D%7B2L%7D)
![f_2 =2\times \dfrac{50}{2\times 2}](https://tex.z-dn.net/?f=f_2%20%3D2%5Ctimes%20%5Cdfrac%7B50%7D%7B2%5Ctimes%202%7D)
f₂= 25 Hz
n = 3
![f = n\dfrac{v}{2L}](https://tex.z-dn.net/?f=f%20%3D%20n%5Cdfrac%7Bv%7D%7B2L%7D)
![f_3 =3\times \dfrac{50}{2\times 2}](https://tex.z-dn.net/?f=f_3%20%3D3%5Ctimes%20%5Cdfrac%7B50%7D%7B2%5Ctimes%202%7D)
f₃ = 37.5 Hz
Answer:
how much the atom weights
Explanation:
Use this equation
Acceleration = frequency x wavelength
You have the wavelength and frequency so multiply them.
3.0 x .5 = 1.5
I don't remeber if any conversions need to be made
Answer:
Explanation:
Radius of circular path of coin R = 12.5 x 10⁻²,
coefficient of static friction μs = .33
In order that the coin rotates in circular path , it requires centripetal force which is provided by friction. As speed of rotation increases , force of friction also increases to provide it required centripetal force. When the speed of rotation becomes too high so that frictional force can not compensate the increase in centripetal force then coin will start slipping or it starts moving with respect to turntable.
b ) At this point of time
centripetal force = limiting force of friction
mω² R = μs mg , m is mass of the coin , ω is angular velocity ,
ω² R = μs g
ω² x 12.5 x 10⁻², = .33 x 9.8
ω² = .33 x 9.8 / 12.5 x 10⁻²,
= 25.87
ω = 5.08 rad / s
2π / T = 5.08
T = 2π / 5.08
= 1.23 s .
Y e s i t ' s c a l l e d " t h e e f f o r t f o r c e "
H o p e t h i s h e l p s
:) L O L