56. What is the range of magnitudes that can be attained from the
1 answer:
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Answer:
Net forces which pushes the window is 30342.78 N.
Explanation:
Given:
Dimension of the office window.
Length of the window = m
Width of the window = m
Area of the window =
Difference in air pressure = Inside pressure - Outside pressure
= atm =
atm
Conversion of the pressure in its SI unit.
⇒ atm =
Pa
⇒ atm =
Pa
We have to find the net force.
We know,
⇒ Pressure = Force/Area
⇒
⇒
⇒ Plugging the values.
⇒
⇒ Newton (N)
So,
The net forces which pushes the window is 30342.78 N.
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