The distance between two slits is 1.50 *10-5 m. A beam of coherent light of wavelength 600 nm illuminates these slits, and the d
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Answer:
Part A) Electric fields at the point due to q₁ and q₂:
E₁ = 33.75*10³ N/C (-j) , E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C
Part B) Net electric field at P (Ep)
Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Equivalence
1nC= 10⁻⁹C
1cm= 10⁻²m
Data
k= 9*10⁹ N*m²/C²
q₁ = -6.00 nC = -6 *10⁻⁹C
q₂ = +3.00 nC = +3*10⁻⁹C
d₁ = 4cm = 4 *10⁻²m
d₂ = 5 *10⁻²m
Part A) Calculation of the electric fields at the point due to q₁ and q₂
Look at the attached graphic:
E₁: Electric Field at point P(0,4) cm due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge
E₂: Electric Field at point P(0,4) cm due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge
E₁ = k*q₁/d₁² = 9*10⁹ *6 *10⁻⁹/ (4 *10⁻²)² = 33.75*10³ N/C
E₂ = k*q₂/d₂²= 9*10⁹ *3*10⁻⁹/(5 *10⁻²)² = 10.8*10³ N/C
E₁ = 33.75*10³ N/C (-j)
E₂x=E₂cosβ = 10.8*(3/5) = 6.48*10³ N/C
E₂y=E₂sinβ = 10.8*(4/5) = 8.64*10³ N/C
E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C
Part B) Calculation of the net electric field at P (Ep)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Ep=Epx (i) + Epy (j)
Epx= E₂x= 6.48*10³ N/C (-i)
Epy= E₁y+E₂y= (33.75*10³ (-j) + 8.64*10³ (+j) ) N/C=25.11 10³ (-j) N/C
Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C
Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C
1) Yes
2)
Explanation:
1)
To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.
The pressure exerted by a column of fluid of height h is:
where
is the atmospheric pressure
is the fluid density
is the acceleration due to gravity
h is the height of the column of fluid
Here we have:
is the sea water density
h = 5440 m is the depth at which the bathyscaphe is located
Therefore, the pressure on it is
Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.
2)
Here we want to find the force exerted on the bathyscaphe.
The relationship between force and pressure on a surface is:
where
p is hte pressure
F is the force
A is the area of the surface
Here we have:
is the pressure exerted
The bathyscaphe has a spherical surface of radius
r = 3 m
So its surface is:
Therefore, we can find the force exerted on it by re-arranging the previous equation:
Answer:
98.15 lb
Explanation:
weight of plane (W) = 5,000 lb
velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s
wing area (A) = 200 ft^{2}
aspect ratio (AR) = 8.5
Oswald efficiency factor (E) = 0.93
density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}
Drag = 0.5 x ρ x x A x Cd
we need to get the drag coefficient (Cd) before we can solve for the drag
Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)
where
- induced drag coefficient (Cdi) =
(take note that π is shown as n and ρ is shown as
)
where lift coefficient (Cl)= =
= 0.245
therefore
induced drag coefficient (Cdi) = =
= 0.0024
- since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)
- Cd = 0.0024 + 0.0024 = 0.0048
Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.
Drag = 0.5 x ρ x x A x Cd
Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb