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:
b) vary with the frequency of the light
Explanation:
The phone electric effect can be expressed as
K.E=(hv -W•)
Where K.E is the Kinectic energy
W• = work function of the metal
ν =frequency of the radiation
h = Planck's constat
Then, we can see that K.E is proportional linearly to "v" in the equation above.
Therefore, When light is directed on a metal surface, the kinetic energies of the photoelectrons vary with the frequency of the light
Answer:
Explanation:
In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:
In our problem, the work function of cesium is
so, we can convert it into Joules by using the following proportion:
Answer:
V = 0.0723 volts = 72.3 milivolts
Explanation:
The emf induced in the rod is the motional emf due to the magnetic field. This motional emf can be calculated by the following formula:
where,
V = Motional EMF = ?
v = speed of rod = 12.5 m/s
B = Magnetic Field = 6.23 mT = 0.00623 T
l = Length of rod = 92.9 cm = 0.929 m
θ = angle between v and B = 90°
Therefore,
<u>V = 0.0723 volts = 72.3 milivolts</u>
Answer:
Explanation:
Given,
initial angular speed, ω = 3,700 rev/min
= 
final angular speed = 0 rad/s
Number of time it rotates= 46 times
angular displacement, θ = 2π x 46 = 92 π
Angular acceleration
