Answer:
A. 1.64 J
Explanation:
First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:
![n=\frac{m}{M_m}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7Bm%7D%7BM_m%7D)
where
n is the number of moles
m = 1.4 mg = 0.0014 g is the mass of mercury
Mm = 200.6 g/mol is the molar mass of mercury
Substituting, we find
![n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B0.0014%20g%7D%7B200.6%20g%2Fmol%7D%3D7.0%5Ccdot%2010%5E%7B-6%7D%20mol)
Now we have to find the number of atoms contained in this sample of mercury, which is given by:
![N=n N_A](https://tex.z-dn.net/?f=N%3Dn%20N_A)
where
n is the number of moles
is the Avogadro number
Substituting,
atoms
The energy emitted by each atom (the energy of one photon) is
![E_1 = \frac{hc}{\lambda}](https://tex.z-dn.net/?f=E_1%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
where
h is the Planck constant
c is the speed of light
is the wavelength
Substituting,
![E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J](https://tex.z-dn.net/?f=E_1%20%3D%20%5Cfrac%7B%286.63%5Ccdot%2010%5E%7B-34%7D%20Js%29%283%5Ccdot%2010%5E8%20m%2Fs%29%7D%7B5.08%5Ccdot%2010%5E%7B-7%7D%20m%7D%3D3.92%5Ccdot%2010%5E%7B-19%7D%20J)
And so, the total energy emitted by the sample is
![E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J](https://tex.z-dn.net/?f=E%3DnE_1%20%3D%20%284.22%5Ccdot%2010%5E%7B18%7D%20%29%283.92%5Ccdot%2010%5E%7B-19%7DJ%29%3D1.64%20J)
It is an imaginary transformer which has no core loss, no ohmic resistance and no leakage flux. The ideal transformer has the following important characteristic. The resistance of their primary and secondary winding becomes zero. The core of the ideal transformer has infinite permeability.
Answer: ![6.408(10)^{-19} C](https://tex.z-dn.net/?f=6.408%2810%29%5E%7B-19%7D%20C)
Explanation:
This problem can be solved by the following equation:
![\Delta K=q V](https://tex.z-dn.net/?f=%5CDelta%20K%3Dq%20V)
Where:
is the change in kinetic energy
is the electric potential difference
is the electric charge
Finding
:
![q=\frac{\Delta K}{V}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B%5CDelta%20K%7D%7BV%7D)
![q=\frac{7.37(10)^{-17} J}{115 V}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B7.37%2810%29%5E%7B-17%7D%20J%7D%7B115%20V%7D)
Finally:
![q=6.408(10)^{-19} C](https://tex.z-dn.net/?f=q%3D6.408%2810%29%5E%7B-19%7D%20C)
I'm pretty sure its d. because depends what tipe of dog it is there are some that see different colors so I'm pretty sure its d.
Answer:
<h2>The angular velocity just after collision is given as</h2><h2>
![\omega = 0.23 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%200.23%20rad%2Fs)
</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>
Explanation:
As per given figure we know that there is no external torque about hinge point on the system of given mass
So here we will have
![L_i = L_f](https://tex.z-dn.net/?f=L_i%20%3D%20L_f)
now we can say
![m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega](https://tex.z-dn.net/?f=m_1v_1%5Cfrac%7BL%7D%7B2%7D%20%3D%20%28m_2L%5E2%20%2B%20m_1%28%5Cfrac%7BL%7D%7B2%7D%29%5E2%29%5Comega)
so we will have
![0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega](https://tex.z-dn.net/?f=0.49%281.89%29%280.45%29%20%3D%20%282.13%280.90%29%5E2%20%2B%200.49%280.45%29%5E2%29%5Comega)
![\omega = 0.23 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%200.23%20rad%2Fs)
Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass
So we can use angular momentum conservation about the hinge point