In meters per second, the singer is traveling ...
... 83 km/h × (1 h)/(3600 s) × (1000 m)/(1 km) ≈ 23 1/18 m/s
Then the observed frequency is ...
... f' = 343/(343 -23.056)·700 Hz ≈ 750.4 Hz
Answer:
0.01j
Explanation:
the energy equals the work done by the bulb.
Workdone=
![workdone = \frac{power}{time}](https://tex.z-dn.net/?f=workdone%20%3D%20%20%5Cfrac%7Bpower%7D%7Btime%7D%20%20)
power=voltage×current
=9×0.2
=1.8 W.
THEREFORE,
time=3×60
= 180s
workdone=1.8/180
=0.01 j
The answer is A. Mixtures can be separated by physical means.
Mixtures can be separated through physical means while pure substances cannot be separated through physical means. Pure substances form compounds which are chemically bonded and can be separated through chemical means.
There are huge losses in the transmission, production and usage of electricity and the reduction of these losses in order to save electricity is called as conservation of energy.
As per the statistics, there is loss of nearly 4% while the transmission of electricity. Like wise during production also, lot of electricity get wasted due to the inefficient material used. None of the production material nor the equipment used have 100% efficiency and thus there is always a possibility of energy wastage.
When it is said that the energy is wasted , it simply means that the energy production which should have been 100% as per calculation is not completely derived from the source due to the inefficient conversion process. For example, a turbine while rotating must convert 100 % of the water energy or water falling on it into electrical energy but the turbine is not able to do so as some of the water is lost or its energy is lost before conversion while going through the mechanical process.
Answer:
(a) 1.716 MeV
(b) 0.284 MeV
(c) ![45^{\circ}](https://tex.z-dn.net/?f=45%5E%7B%5Ccirc%7D)
Solution:
As per the question:
Energy of the scattered neutron, E = 2 MeV = ![2\times 10^{6}\ eV](https://tex.z-dn.net/?f=2%5Ctimes%2010%5E%7B6%7D%5C%20eV)
Atomic no. of ![^{12}\textrm{C},\ A = 12](https://tex.z-dn.net/?f=%5E%7B12%7D%5Ctextrm%7BC%7D%2C%5C%20A%20%3D%2012)
Now,
To calculate the Kinetic Energy of the scattered neutron, ![E_{av} = \frac{1}{2}(1 - \alpha)E](https://tex.z-dn.net/?f=E_%7Bav%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%281%20-%20%5Calpha%29E)
where
![\alpha = (\frac{A - 1}{A + 1})^{2}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%28%5Cfrac%7BA%20-%201%7D%7BA%20%2B%201%7D%29%5E%7B2%7D)
Therefore,
![E_{av} = \frac{1}{2}(1 - (\frac{A - 1}{A + 1})^{2})E](https://tex.z-dn.net/?f=E_%7Bav%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%281%20-%20%28%5Cfrac%7BA%20-%201%7D%7BA%20%2B%201%7D%29%5E%7B2%7D%29E)
![E_{av} = \frac{1}{2}(1 - (\frac{12 - 1}{12 + 1})^{2})\times 2\times 10^{6} = 0.284\ MeV](https://tex.z-dn.net/?f=E_%7Bav%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%281%20-%20%28%5Cfrac%7B12%20-%201%7D%7B12%20%2B%201%7D%29%5E%7B2%7D%29%5Ctimes%202%5Ctimes%2010%5E%7B6%7D%20%3D%200.284%5C%20MeV)
(a) The energy of the scattered neutron is: 2 MeV - 0.284 MeV = 1.716 MeV
(b) The recoil energy of the
nucleus is 0.284 MeV
(c) The angle of the recoil of the nucleus is ![45^{\circ}](https://tex.z-dn.net/?f=45%5E%7B%5Ccirc%7D)