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:
a. b- x= y
dx = -dy
b. F = 
c. F = 
Explanation:
a. x components:
= 
Integrating and solving gives:
b- x= y
dx = -dy
b. the force is given by the equation derived from (a.):
F =
c. Given that r>>a, the expression becomes:
F =
Explanation:
When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.
Answer:
Mass, m = 125 kg
Explanation:
Let us assume that the question says, "What is the mass of an object whose velocity is 400 m/s and the kinetic energy of 10⁷ J.
The kinetic energy of an object is :
So, the mass of the object is 125 kg.
Answer:
a. 
W
Explanation:
= temperature of the surface of sun = 5800 K
= Radius of the Sun = 7 x 10⁸ m
= Surface area of the Sun
Surface area of the sun is given as
= Emissivity = 1
= Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴
Using Stefan's law, Power output of the sun is given as
Answer:
plug it into your calculator
Explanation:
Kinetic energy = KE = Joules = J
KE= 1/2mv^2
90,000 = 1/2 m (12^2)
180,000 = m (144)
m = 180,000 / 144
