A negative oil droplet is held motionless in a millikan oil drop experiment. What happens if the switch is opened?

Answer:

130 m/s (to two significant figures)

Explanation:

In projectile motion, the launching velocity and launch angle help to determine both the horizontal and vertical velocity components.

u represents the initial projectile velocity = 150 m/s

uₓ = u cos θ = 150 cos 30° = 129.9 m/s

uᵧ = u sin θ = 150 sin 30° = 75.0 m/s

A projectile's motion can be viewed as made up of independent vertical and horizontal elements.

The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.

Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.

Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.

At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s

assuming north-south is along the Y-axis and east-west along the X-axis

X = total X-displacement

from the graph, total displacement in the X-direction is computed as

X = 0 - 20 + 60 Cos45 + 0

X = 42.42 - 20

X = 22.42 m

Y = total Y-displacement

from the graph, total displacement in the Y-direction is computed as

Y = 40 + 0 + 60 Sin45 + 50

Y = 90 + 42.42

Y = 132.42 m

To calculate the magnitude of the net displacement vector, we apply the Pythagorean theorem, yielding

magnitude: Sqrt(X² + Y²) = Sqrt(22.42² + 132.42²) = 134.31 m

Direction: tan⁻¹(Y/X) = tan⁻¹(132.42/22.42) = 80.4 deg north of east

Answer with Explanation:

Concepts and reasoning

The principle for addressing this question is that a capacitor in an RC circuit allows current to flow until fully charged. Once charged, it prevents any further current from moving through. Conversely, the situation is different with an inductor in an RL circuit. In accordance with Faraday's law, an inductor generates an electromagnetic force to counteract the applied voltage, but when no change in flux occurs, it behaves akin to a regular wire as if the inductor is absent.

In the accompanying diagram, a resistor is connected in series with a capacitor.

As we observe

the voltage across both the capacitor and the source.

Voltage across a resistor in an RC circuit.

Voltage across a resistor in an RL circuit.

B) 14.0 N

To address this inquiry, we need to evaluate the kinetic energy of the box before and after crossing the rough section. The kinetic energy is given by the formula:

E = 0.5 M V^2

where

E = Energy

M = Mass

V = velocity

Now, utilizing the known data, we compute the energy prior and post.

Before:

E = 0.5 M V^2

E = 0.5 * 13.5kg * (2.25 m/s)^2

E = 6.75 kg * 5.0625 m^2/s^2

E = 34.17188 kg*m^2/s^2 = 34.17188 joules

After:

E = 0.5 M V^2

E = 0.5 * 13.5kg * (1.2 m/s)^2

E = 6.75 kg * 1.44 m^2/s^2

E = 9.72 kg*m^2/s^2 = 9.72 Joules

Hence, the box consumed energy equal to 34.17188 J - 9.72 J = 24.451875 J over a length of 1.75 meters. Next, we will calculate the loss per meter by dividing the energy loss by the distance traversed.
24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N

When we round to one decimal point, we arrive at 14.0 N, which corresponds with option “B.”

Answer:

a) Ф = 0.016 N / C m, b) q_{int} = 0.14 10⁻¹² C

Explanation:

a) For this scenario, we rely on Gauss's law

          Ф = E.ds = /ε₀

As the field points in the x direction, there is no flux through the cylinder walls.

          Ф = E A

         

The area of a circle is

           A = π r

          Ф = E π r

          Ф = (x- 3.6) r

Now, let's compute

          Ф = (3.7 -3.6) 0.16

          Ф = 0.016 N / C m

     

b) Using Gauss's law, we have

             q_{int} = Ф ε₀

 

Where the flow is present on both sides, at the face corresponding to x = 0, the flow is zero

             q_{int} = 0.016 8.85 10⁻¹²

             q_{int} = 0.14 10⁻¹² C