Answer:
The energy of photon, 
Explanation:
It is given that,
Voltage of anode, 
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 is charge of electron


So, the maximum energy of the x- ray radiation is
. Hence, this is the required solution.
The answer to that question is c. tamod
It must be sliding friction, because the fish is already in motion.
Kinetic energy has nothing to do with anything other than motion of the particle.
When a particle with velocity v collides another particle(suppose it is at rest for simplication), assuming that there is perfectly elastic collision between them, the velocity of particle which was at rest becomes mv/M ( assuming mass of particle in motion to be m and at rest to be M) from convervation of linear momentum. And all this transfer of energy happens in a fraction of seconds which is not visible to naked eyes.
Hence 1st option is correct!
Answer:
F=1.65 x 10²⁶ N
Explanation:
Given that
Distance ,R= 3.34 x 10¹² m
Mass m₁= 2.78 x 10³⁰ kg
Mass ,m₂= 9.94 x 10³⁰ kg
we know that gravitational force F given as

G=Constant
G=6.67 x 10⁻¹¹ Nm²/kg²
Now by putting the values

F=1.65 x 10²⁶ N
Therefore the force between these two mass will be 1.65 x 10²⁶ N.