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
Necrosis - the death of most or all of the cells in an organ or tissue due to disease, injury, or failure of the blood supply.
Hope this helps, please mark me brainliest
Answer:
The last graph.
Explanation:
Gravitational potential energy is the energy possessed by a body at a given height from the Earth's surface.
The formula to find the gravitational potential energy is given as:
Where, 'U' is the gravitational potential energy.
'm' is the mass of the body.
'g' is the acceleration of the body due to gravity.
'h' is the height of the body above the Earth's surface.
So, from the above equation, it is clear that, gravitational potential energy is directly proportional to the height. So, as height increases, the gravitational potential energy increases. At the surface of Earth, where, height is 0, the gravitational potential energy is also zero.
Therefore, the correct graph is a straight line with positive slope and passing through the origin. So, the last option is the correct one.
Answer:
I'm sorry, I don't think there is any answer to give seeing as no question has been asked
