<h2>Question:</h2>
In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?
Answer:
9.1Ω
Explanation:
The circuit diagram has been attached to this response.
(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e
=> 
------------(i)
From the question;
R1 = 3Ω,
R2 = 7Ω
Substitute these values into equation (i) as follows;
Ω
(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.
Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e
R = Rₓ + R3
Rₓ = 2.1Ω
R3 = 7Ω
=> R = 2.1Ω + 7Ω = 9.1Ω
Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω
C, velocity I believe this is the answer to your question
Answer:
Explanation:
Equal to the sum of the densities of both substances
Equal to the sum of the densities of both substances
Equal to the difference in the densities of both substances
Equal to the difference in the densities of both substances
of which I cannot predict anything about its value
of which I cannot predict anything about its value
of an intermediate value to both substances which will depend on which substance is in greater quantity
of an intermediate value to both substances which will depend on which substance is in greater quantity
The velocity of the second glider after the collision is 4.33 m/s rightward.
<h3>
Velocity of the second glider after the collision</h3>
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
- m₁ is mass of first glider
- m₂ is mass of second glider
- u₁ is initial velocity of first glider
- u₂ is initial velocity of second glider
- v is the final velocity of the gliders
(2)(1) + (3)(5) = (2)(2) + 3v₂
17 = 4 + 3v₂
3v₂ = 17 - 4
3v₂ = 13
v₂ = 13/3
v₂ = 4.33 m/s
Thus, the velocity of the second glider after the collision is 4.33 m/s rightward.
Learn more about linear momentum here: brainly.com/question/7538238
#SPJ1
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
