How can a calculated height be greater than an actual height?

A rod bent into the arc of a circle subtends an angle 2θ at the center P of the circle (see below). If the rod is charged unifor

Zigmanuir [339]

Answer:

Qsinθ/4πε₀R²θ

Explanation:

Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.

Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.

So dq = λRdθ.

Substituting dq into dE', we have

dE' = dqcosθ/4πε₀R²

= λRdθcosθ/4πε₀R²

= λdθcosθ/4πε₀R

E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ

E' = (λ/4πε₀R)[sinθ] from -θ to θ

E' = (λ/4πε₀R)[sinθ]

= (λ/4πε₀R)[sinθ - sin(-θ)]

= (λ/4πε₀R)[sinθ + sinθ]

= 2(λ/4πε₀R)sinθ

= (λ/2πε₀R)sinθ

Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ

Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ

Q = 2λRθ

λ = Q/2Rθ

Substituting λ into E', we have

E' = (Q/2Rθ/2πε₀R)sinθ

E' = (Q/θ4πε₀R²)sinθ

E' = Qsinθ/4πε₀R²θ where θ is in radians

5 0

3 years ago

A surface is bombarded by particles, each of mass small 'm', which have velocity

ale4655 [162]

Answer:

B. 2nmv

Explanation:

Pressure is force over area.

P = F / A

Force is mass times acceleration.

F = ma

Acceleration is change in velocity over change in time.

a = Δv / Δt

Therefore:

F = m Δv / Δt

P = m Δv / (A Δt)

The total mass is nm.

The change in velocity is Δv = v − (-v) = 2v.

A = 1 and Δt = 1.

Plugging in:

P = (nm) (2v) / (1 × 1)

P = 2nmv

8 0

3 years ago

How does a parallel circuit differ from a series circuit?

8_murik_8 [283]

I think the correct answer from the choices listed above is option B. A parallel circuit differ from a series circuit in a sense that a <span>series circuit has one path for electrons, but a parallel circuit has more than one path. In a parallel circuit there two or more paths for current to flow while a series circuit only has one.</span>

5 0

3 years ago

A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc

Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0

3 years ago

aleksklad [387]

1. Mitosis and Meiosis are both ways cells duplicate DNA
2. They both go through PMAT ( prophase, metaphase, anaphase and telophase)

8 0

3 years ago