Answer:
man will move in opposite direction with speed

Explanation:
As we know that man is lying on the friction-less surface
so here net force along the surface is zero
so if we take man + stone as a system then net change in momentum of this system will become zero
so here we have


here we have



<span>Pice=920kg/m^3
deltaP=PgH=920kg/m^3 X 9.80665m/s^2 X 1000m = 9022118 Pa
P=Po + deltaP=101.325 + 9022 = 9123kPa</span>
Explanation:
It is given that,
Magnetic field, B = 0.1 T
Acceleration, 
Charge on electron,
Mass of electron,
(a) The force acting on the electron when it is accelerated is, F = ma
The force acting on the electron when it is in magnetic field, 
Here, 
So, 
Where
v is the velocity of the electron
B is the magnetic field


v = 341250 m/s
or

So, the speed of the electron is 
(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.
'Intelligent person' could be a possibility for your answer, also a 'scientist' or a 'philosopher' as well as 'an old person' may equal to the meaning of being 'wise humans'.
Answer:
197.5072.
Explanation:
According to the Coulomb's law, the magnitude of the electrostatic force of interaction between two charges
and
which are separated by the distance
is given by

<em>where,</em> k is the Coulomb's constant.
For the case, when,
Then, using Coulomb's law,

For the case, when,
Then, using Coulomb's law, the new electric force between the charges is given by,
