Answer:
The right solution is "24.39 per sec".
Explanation:
According to the question,
⇒ ![v=\frac{502.1}{\sqrt{3} }](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B502.1%7D%7B%5Csqrt%7B3%7D%20%7D)
![=289.9 \ m/s](https://tex.z-dn.net/?f=%3D289.9%20%5C%20m%2Fs)
The time will be:
⇒ ![t=\frac{d}{v}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7Bv%7D)
![=\frac{2\times 6}{289.9}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%5Ctimes%206%7D%7B289.9%7D)
![=\frac{12}{289.9}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B12%7D%7B289.9%7D)
![=0.041 \ sec](https://tex.z-dn.net/?f=%3D0.041%20%5C%20sec)
hence,
⇒ ![N=\frac{1}{t}](https://tex.z-dn.net/?f=N%3D%5Cfrac%7B1%7D%7Bt%7D)
![=\frac{1}{0.041}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B0.041%7D)
Answer:
The wavelength of the wave is 20 m.
Explanation:
Given that,
Amplitude = 10 cm
Radial frequency ![\omega = 20\pi\ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%2020%5Cpi%5C%20rad%2Fs)
Bulk modulus = 40 MPa
Density = 1000 kg/m³
We need to calculate the velocity of the wave in the medium
Using formula of velocity
![v=\sqrt{\dfrac{k}{\rho}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cdfrac%7Bk%7D%7B%5Crho%7D%7D)
Put the value into the formula
![v=\sqrt{\dfrac{40\times10^{6}}{10^3}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cdfrac%7B40%5Ctimes10%5E%7B6%7D%7D%7B10%5E3%7D%7D)
![v=200\ m/s](https://tex.z-dn.net/?f=v%3D200%5C%20m%2Fs)
We need to calculate the wavelength
Using formula of wavelength
![\lambda =\dfrac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cdfrac%7Bv%7D%7Bf%7D)
![\lambda=\dfrac{v\times2\pi}{\omega}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bv%5Ctimes2%5Cpi%7D%7B%5Comega%7D)
Put the value into the formula
![\lambda=\dfrac{200\times2\pi}{20\pi}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7B200%5Ctimes2%5Cpi%7D%7B20%5Cpi%7D)
![\lambda=20\ m](https://tex.z-dn.net/?f=%5Clambda%3D20%5C%20m)
Hence, The wavelength of the wave is 20 m.
There are two conditions necessary for total internal reflection, which is when light hits the boundary between two mediums and reflects back into its original medium:
Light is about to pass from a more optically dense medium (slower) to a less optically dense medium (faster).
The angle of incidence is greater than the defined critical angle for the two mediums, which is given by:
θ = sin⁻¹(
/
)
Where θ = critical angle,
= refractive index of faster medium,
= refractive index of slower medium.
Choice C gives one of the above necessary conditions.
A. 90km/h = 25 m/s
W=1/2mv^2
=1/2* 650* 25^2
= 203125(J)
b. v’= 90-50= 40km/h
= 100/9 m/s
W=1/2mv^2
=1/2*650*(100/9)^2
= 40123,45679
I found: 16,905J
Explanation:
The Gravitational Potential Energy will depend on the work done against gravity (weight,
W=mg) to lift it at height h or:
U = mgh =750 * 9.8 *2.3 =16,905J