<u>Given </u><u>:</u><u>-</u>
- An elevator is moving vertically up with an acceleration a.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- The force exerted on the floor by a passenger of mass m .
<u>Solution</u><u> </u><u>:</u><u>-</u>
As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
For the FBD refer to the attachment . From that ,
<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>
Answer:
D by spinning magnets to induce electric current.
Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA = 
/ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper
The cold air sinks as it is heavier
