Answer:
Value of electric field along the axis and equitorial axis and
respectively.
Explanation:
Given :
Distance between charges ,
Magnitude of charges ,
Dipole moment ,
Case A) (x,y) = (12.0 cm, 0 cm) :
Electric field of dipole in its axis ,
Putting all values and
We get ,
Case B) (x,y) = (0 cm, 12.0 cm) :
Electric field of dipole on equitorial axis ,
Putting all values and
We get ,
Hence , this is the required solution.
Answer:
<em>The force required is 3,104 N</em>
Explanation:
<u>Force</u>
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
F = ma
Where a is the acceleration of the object.
On the other hand, the equations of the Kinematics describe the motion of the object by the equation:
Where:
vf is the final speed
vo is the initial speed
a is the acceleration
t is the time
Solving for a:
We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:
The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:
F= 3,104 N
The force required is 3,104 N
Answer:
Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e ()
with this expression we can find the closest approach distance (r)