All categories

2 years ago

How do we solve questions C and D? I already did A and B and I am confused on how to continue

Physics

1 answer:

aleksandrvk [35]2 years ago

8 0

(a) The work done in moving the unit charge from point C to A is 7.62 x 10⁻³ J.

(b) The work done in moving the unit charge from point D to B is 7.62 x 10⁻³ J.

<h3>Work done in moving the charge from C to A</h3>

W = Fd

W = Kq²/d

from 0 origin to C, d = √(5² + 5²) = 7.07 m
from 0 origin to A, d = 5 m

W(C to A) = W(0 to C) + W(0 to A)

$W(C \ to \ A) = - \frac{Kq^2}{7.07} + \frac{Kq^2}{5} \\\\ W(C \ to \ A) = 0.0586 \ Kq^2\\\\W(C \ to \ A) = 0.0586 \times 9 \times 10^9 \times (3.8\times 10^{-6})^2\\\\W(C \ to \ A) = 7.62 \times 10^{-3} \ J$

<h3>Work done in moving the charge from D to B</h3>

from 0 origin to D, d = √(5² + 5²) = 7.07 m
from 0 origin to B, d = 5 m

W(D to B) = W(0 to D) + W(0 to B)

W(D to B) = 7.62 x 10⁻³ J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

You might be interested in

A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of th

pentagon [3]

Complete question is;

A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of the conductors acing between the centers of the conductors (say, a and b) is 2.5 m. A telephone line is also symmetrically supported on a horizontal cross arm 1.8 m directly below the power line. Spacing between the centers of these conductors (say, c and d) is 1.0 m.

The mutual inductance per unit length between circuit a-b and circuit c-d is given as 4 x 10^(-7) ln √((D_ad × D_bc)/(D_ac × D_bd)) H/m

where, for example, D_ad denotes the distance in meters between conductors a and d.

a. Hence, compute the mutual inductance per kilometer between the power line and the telephone line.

b. Find the 60-Hz voltage per kilometer induced in the telephone line when the power line carries 150 A

Answer:

A) M = 1.01 × 10^(-4) H/km

B) v_cd = 5.712 V/km

Explanation:

A) From the distances given in the question, we can deduce that;

D_ac = √(((2.5/2) - (1/2))² + 1.8²)

D_ac = 1.95 m

Also;

D_ad = √(((2.5/2) + (1/2))² + 1.8²)

D_ad = 2.51 m

I_a and I_b are put of phase by 180°. Thus, due to a and b, the flux linkages to c and d is given as;

φ_cd = 4 x 10^(-7)I_a( ln (2.51/1.95))

Mutual inductance per km is given as;

M = φ_cd/I_a

Thus;

M = 4 x 10^(-7)( ln (2.51/1.95))

M = 1.01 × 10^(-7) H/m

Per km;

M = 1.01 × 10^(-7) × 1000

M = 1.01 × 10^(-4) H/km

B) voltage per km is gotten by;

v_cd = ωMI

Now, ω = 2πf = 2π × 60 = 377 rad/s

Thus;

v_cd = 377 × 1.01 × 10^(-4) × 150

v_cd = 5.712 V/km

5 0

2 years ago

while standing on the sidewalk facing the road, you see a bicyclist passing by toward your right. in the reference frame of the

cricket20 [7]

Answer:

Consider frames X and Y:

If X sees Y moving to his right then Y must see X moving to his right.

If this is not true then one can choose one frame over the other ( a favored frame and this is not allowed)

7 0

3 years ago

g what are the bands of electromagneitc radiation and how are they related to energy frequency and wavelength

ch4aika [34]

Answer:

The details of bands is given in explanation.

Explanation:

The electromagnetic waves are differentiated into different bands based upon their wavelengths and frequencies. The names of different bands are as follows:

1. Radio Waves

2. Micro Waves

3. Infra-red

4. Visible light

5. Ultra Violet

6. X-rays

7. Gamma Rays

<u>The frequency of every region or rays increases from 1 through 7</u>. The <u>energy of rays also increase from 1 through 7</u>. Since, the wave length is inversely related to energy and frequency, thus the <u>wavelength of rays decrease from 1 through 7</u>.

A detailed information of the bands is provided in the picture attached.

3 0

3 years ago

The question is on the picture

Firlakuza [10]

Answer:

<em> think 2 also if not im so sorry but i think it is :)</em>

7 0

2 years ago

A motion sensor emits sound, and detects an echo 0.0115 s after. A short time later, it again emits a sound, and hears an echo a

Mekhanik [1.2K]

Answer:

1.17 m

Explanation:

From the question,

s₁ = vt₁/2................ Equation 1

Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.

Given: v = 343 m/s, t = 0.0115 s

Substitute into equation 1

s₁ = (343×0.0115)/2

s₁ = 1.97 m.

Similarly,

s₂ = vt₂/2.................. Equation 2

Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo

Given: v = 343 m/s, t₂ = 0.0183 s

Substitute into equation 2

s₂ = (343×0.0183)/2

s₂ = 3.14 m

The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁

s₂-s₁ = (3.14-1.97) m = 1.17 m

7 0

3 years ago

Read 2 more answers