2. Is there a proportional relationship between voltage and current? If so, write the equation for each run in the form potentia
1 answer:
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a) See free-body diagram in attachment
b) The acceleration is
Explanation:
a)
The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.
The free-body diagram for this object is shown in the figure in attachment.
There are three forces acting on the object:
- The weight of the object, labelled as
(where m is the mass of the object and g is the acceleration of gravity), acting downward
- The applied force,
, acting up along the plane
- The force of friction,
, acting down along the plane
b)
In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:
where:
is the applied force, pushing forward
is the frictional force, acting backward
is the component of the weight parallel to the incline, acting backward, where
m = 2 kg is the mass of the object
is the acceleration of gravity
is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)
a is the acceleration
Solving for a, we find:
Learn more about inclined planes:
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Answer:
q₁ = +12.19 μC
q₂ = +55.81 μC
Explanation:
From Coulomb's law,
Where,
F is the electrostatic force
k is the constant of proportionality 9 × 10⁹ Nm²/C²
q₁ and q₂ are the point charges
r is the distance between the charges
q₁q₂ = 6.804 × 10 ⁻¹⁰ C
q₁q₂ = 0.6804 × 10 ⁻⁹ C -------(1)
Also, the total charge of the system is 68.0 μC
⇒ q₁ + q₂ = 68.0 μC
For simplicity, q₁ + q₂ = 68 × 10⁻⁶ C --------(2)
Solving equation 1 and 2 simultaneously using substitution method. From equation 2,
q₂ = 68 × 10⁻⁶ - q₁ ---------(3)
Substituting equation (3) into (1)
q₁(68 × 10⁻⁶ - q₁) = 0.6804 × 10 ⁻⁹
68 × 10⁻⁶q₁ - q₁² = 0.6804 × 10 ⁻⁹
q₁² - 68 × 10⁻⁶q₁ + 0.6804 × 10 ⁻⁹ = 0
solving the quadratic equation using quadratic formula,
a = 1, b = - 68 × 10⁻⁶, c = 0.6804 × 10 ⁻⁹
q₁ = +12.19 μC or +55.81 μC
Testing for the two q₁ to get q₂
When q₁ = +12.19 μC
12.19 μC + q₂ = 68.0 μC
q₂ = 68.0 μC - 12.19 μC
q₂ = +55.81 μC
When q₁ = +55.81 μC
55.81 μC + q₂ = 68.0 μC
q₂ = 68.0 μC - 55.81 μC
q₂ = +12.19 μC
We have two possible solutions for the question.
When q₁ = +12.19 μC, q₂ = +55.81 μC and
When q₁ = +55.81 μC, q₂ = +12.19 μC
Considering the conditions given in the system,
(1) Two positive point charges q₁ and q₂
(2) Charge q₂ is greater than charge q₁, q₂>q₁
The solution that satisfies the condition is q₁ = +12.19 μC, q₂ = +55.81 μC