When a pendulum is pulled back from its equilibrium position by 10∘, the restoring force is 1.0 N. When it is pulled back to 30∘

What's the response shshshsbshsbsbs sbo

Since you've completed parts a and b, I will tackle part c.
For part C
To respond to this question, we must identify the zeros of the velocity function:

This polynomial can be factored:

Finding the zeros now becomes straightforward since the function equals zero when any factor is zero.

By solving these equations, we identify our zeros:

The particle remains stationary at t=0 and t=3/2.
For part D
We must discover when the velocity function exceeds zero. We will utilize its factored form.
We will assess when each factor is greater than zero and compile the findings in the following table:

From the table, it's evident that our function is positive when and t>3/2.
This indicates the interval during which the particle moves forward.
For part E
The distance traveled can be represented as:

We simply substitute t=12 to calculate the total distance traveled:

For part F
Acceleration is defined as the rate at which velocity changes.
We determine acceleration by deriving the velocity function concerning time.

To find the acceleration at 1 second, we substitute t=1s into the previous equation:

Answer:

A) and B) are valid.

Explanation:

When an object remains at rest, it is indicative that no net force acts upon it.

The downward gravitational force from Earth must be counterbalanced by an upward force of equal magnitude in order to maintain rest.

This upward force is provided by the normal force, which adjusts to satisfy Newton’s 2nd Law and is always perpendicular to the surface supporting the object (in this instance, the ground).

At the molecular level, this normal force comes from the ground's bonded molecules acting like tiny springs, compressed by the object’s molecules, providing an upward restorative force.

Thus, statements A) and B) are true.

The rod measures 450mm in length, while the disk has a radius of 75mm. An upward-supporting pin holds the assembly in place when Θ=0, and there exists a torsional spring with a constant of k=20N m/rad at the pin. One end of the rod connects to the pin, while the other connects to the disk.

Answer:

Part A) Electric fields at the designated point due to charges q₁ and q₂:

E₁ = 33.75 * 10³ N/C (-j), E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C

Part B) The overall electric field at P (Ep)

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Explanation:

Conceptual analysis

The electric field at point P caused by a point charge is calculated as:

E = k*q/d²

E: Electric field measured in N/C

q: charge magnitude in Newtons (N)

k: electric constant measured in N*m²/C²

d: distance from the charge q to point P in meters (m)

Equivalence:

1 nC = 10⁻⁹ C

1 cm = 10⁻² m

Data:

k = 9 * 10⁹ N*m²/C²

q₁ = -6.00 nC = -6 * 10⁻⁹ C

q₂ = +3.00 nC = +3 * 10⁻⁹ C

d₁ = 4 cm = 4 * 10⁻² m

d₂ = 5 * 10⁻² m

Part A) Calculation for electric fields at point from q₁ and q₂:

Refer to the attached illustration:

E₁: Electric Field at point P(0,4) cm due to charge q₁. Since q₁ is negative (q₁-), the electric field approaches the charge.

E₂: Electric Field at point P(0,4) cm due to charge q₂. Since q₂ is positive (q₂+), the electric field emanates from 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 for net electric field at P (Ep)

The electric field at point P from multiple point charges is the vector sum of the individual electric fields.

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