Answer:
The energy of photon, ![E=8\times 10^{-15}\ J](https://tex.z-dn.net/?f=E%3D8%5Ctimes%2010%5E%7B-15%7D%5C%20J)
Explanation:
It is given that,
Voltage of anode, ![V=50\ kV=50\times 10^3\ V=5\times 10^4\ V](https://tex.z-dn.net/?f=V%3D50%5C%20kV%3D50%5Ctimes%2010%5E3%5C%20V%3D5%5Ctimes%2010%5E4%5C%20V)
We need to find the maximum energy of the photon of the x- ray radiation. The energy required to raise an electron through one volt is called electron volt.
![E=eV](https://tex.z-dn.net/?f=E%3DeV)
e is charge of electron
![E=1.6\times 10^{-19}\times 5\times 10^4](https://tex.z-dn.net/?f=E%3D1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%205%5Ctimes%2010%5E4)
![E=8\times 10^{-15}\ J](https://tex.z-dn.net/?f=E%3D8%5Ctimes%2010%5E%7B-15%7D%5C%20J)
So, the maximum energy of the x- ray radiation is
. Hence, this is the required solution.
Answer:
v = 3.84 m/s
Explanation:
In order for the riders to stay pinned against the inside of the drum the frictional force on them must be equal to the centripetal force:
![Centripetal\ Force = Frictional\ Force\\\\\frac{mv^2}{r} = \mu R = \mu W\\\\\frac{mv^2}{r} = \mu mg\\\\\frac{v^2}{r} = \mu g\\\\v = \sqrt{\mu gr}](https://tex.z-dn.net/?f=Centripetal%5C%20Force%20%3D%20Frictional%5C%20Force%5C%5C%5C%5C%5Cfrac%7Bmv%5E2%7D%7Br%7D%20%3D%20%5Cmu%20R%20%3D%20%5Cmu%20W%5C%5C%5C%5C%5Cfrac%7Bmv%5E2%7D%7Br%7D%20%3D%20%5Cmu%20mg%5C%5C%5C%5C%5Cfrac%7Bv%5E2%7D%7Br%7D%20%3D%20%5Cmu%20g%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cmu%20gr%7D)
where,
v = minimum speed = ?
g = acceleration due to gravity = 9.81 m/s²
r = radius = 10 m
μ = coefficient of friction = 0.15
Therefore,
![v=\sqrt{(0.15)(9.81\ m/s^2)(10\ m)}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%280.15%29%289.81%5C%20m%2Fs%5E2%29%2810%5C%20m%29%7D)
<u>v = 3.84 m/s</u>
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Answer
Inertia is the resistance of any physical object to any change in its velocity. This includes changes to the object's speed, or direction of motion. An aspect of this property is the tendency of objects to keep moving in a straight line at a constant speed, when no forces act upon them.