Answer:
Approximately 
(given that the magnitude of this charge is 
.)
Explanation:
If a charge of magnitude 
is placed in an electric field of magnitude 
, the magnitude of the electrostatic force on that charge would be 
.
The magnitude of this charge is 
. Apply the unit conversion 
:
.
An electric field of magnitude 
would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
Explanation:
The first equation of motion in kinematics is given by :
.....(1)
u is initial speed
a is acceleration
v is final speed
t is time
Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.
Answer:
19500kg
Explanation:
If you divide 195000N by 10
Power is the work performed divided by time in seconds. The work is the force times distance:
the time in seconds is 60*11+30=690s so the power is
So choice B
Answer:
Increases.
Explanation:
The electric potential increases when the two positive charges of same magnitude bring close to one charge to another because there is repulsive force between them due to same charge and when the two opposite charges move away from each other, the potential energy decreases. When two opposite charges are brought closer together, electric potential energy decreases while on the other hand, when we move opposite charges apart from each other than the work done against the attractive force that leads to an increase in electric potential energy.
