Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;
E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C
Answer:
, 
Explanation:
The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.
(1)
Where 
is the final velocity, 
is the initial velocity, 
is the acceleration and 
is the distance traveled.
Equation (1) can be rewritten in terms of ax:
(2)
Since the plane starts from rest, its initial velocity will be zero (
):
Replacing the values given in equation 2, it is gotten:
So, The acceleration of the plane is 
Now that the acceleration is known, the next equation can be used to find out the time:
(3)
Rewritten equation (3) in terms of t:
<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>
Answer:
0 m/s
Explanation:
Velocity is displacement over time. Displacement is the distance between your initial position and your final position. If you walk in a circle, you end up back where you started, so your displacement is 0. Therefore, your velocity is also 0.
