Answer:
Distance 5 km, Displacement 3 km east
Explanation:
Answer:
0.25683
Explanation:
t = Time taken by the truck to stop = 11 seconds
u = Initial velocity of the truck = 62 mi/h
v = Final velocity of the truck = 0
a = Acceleration of the truck
g = Acceleration due to gravity = 9.81 m/s²
Equation of motion


The minimum coefficient of static friction between the turtle and the truck bed surface needed to prevent the turtle from sliding is 0.25683
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
<span>THIS IS A GAS PHASE REACTION AND WE ARE GIVE PARTIAL PRESSURES . I WRITE IN TERMS OF P RATHER THAN CONCENTRATION :
lnPso2cl12=-kt+lnPso2cl1
initial partial pressure Pso2cl12 the rate constant k and the time t
lnPso2cl12=(4.5*10-2*s-1)*65*s+ln (375)
so lnPso2cl12=3.002
we take the base e antilog:
lnPso2cl12=e3.002
Pso2cl12=20 torr
we use the integrated first order rate
lnPso2cl12=3.002=k*t+ lnPso2cl12=3.002
we use the same rate constant and initial pressure
k=4.5*10-2*s-1
Pso2cl12=375
Pso2cl12=1* so2cl12
Pso2cl12=37.5 torr
subtract in Pso2cl12 grom both side
lnPso2cl12- lnPso2cl12=-kt
ln(x)-ln(y)=ln (x/y)
ln (Pso2cl12/Pso2cl20)=-kt
we get t
-1/k*ln(Pso2cl12/Pso2cl20)=t
t=51 s</span>