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3 years ago

Two point charges, 1.8 pC and −1.8 pC, are separated by 7 µm. What is the dipole moment of this pair of charges?

Physics

1 answer:

Salsk061 [2.6K]3 years ago

3 0

Answer: 12,600,000Cm

Explanation:

From the data's;

Charges(q) = 1.8 PC equal to 1.8 x 10^¹²C

Distance = 7 micrometer, is equal to 0.0000070m

From the equation of electric dipole moment, p= q x d, where q= charge, d=distance and p is the dipole moment.

Then we have 1.8x10^¹² x 0.0000070= 12,600,000Cm

NB: The charges are identical.

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Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let |a| = 22 lb and |b| = 16 lb. Rou

SVEN [57.7K]

Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)

Answer:

Resultant Force=33.8 lb

Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

$\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j$

Total Force

$F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j$

The Resultant Force is given as

$|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb$

For(ii) angle

We can find the angle bu using tanα=y/x

$tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}$

7 0

3 years ago

1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.

marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

$B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB$

<h3>Speed of sound at the given temperature</h3>

$v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s$

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

$f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )$

where;

-v₀ is velocity of the observer moving away from the source
-vs is the velocity of the source moving towards the observer
fs is the source frequency
fo is the observed frequency
v is speed of sound

$f_0 = f_s(\frac{v-v_0}{v- v_s} )$

$f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz$

Learn more about intensity of sound here: brainly.com/question/17062836

3 0

2 years ago

How long would it take a machine to do 5.000

Gala2k [10]

Answer:

Explanation:

How long would it take a machine to do 5.000

joules of work if the power rating of the machine

is 100 watts?

6 0

3 years ago

Pls help me… I missed the whole lesson and I have no clue what to do-

galben [10]

Answer:

1) 50 facing towards the right

2) 150 facing right

3) 200 facing right

4) 0- no direction

5) 50- facing left

6) 50 facing right

Explanation:

forces in opposite directions and equal magnitudes counteract each other. in number 2 they face the same direction so they would just be added. in number 4 they oppose each other so would be subtracted

8 0

3 years ago

What grade of sprain is a completely torn ligament?

Alex Ar [27]

Grade 1: Stretching or slight tearing of the ligament with mild tenderness, swelling and stiffness. The ankle feels stable and it is usually possible to walk with minimal pain.

Grade 2: A more severe sprain, but incomplete tear with moderate pain, swelling and bruising. Although it feels somewhat stable, the damaged areas are tender to the touch and walking is painful.

Grade 3: This is a complete tear of the affected ligament(s) with severe swelling and bruising. The ankle is unstable and walking is likely not possible because the ankle gives out and there is intense pain.

source - https://www.rushcopley.com/health/physician-articles/varying-degrees-of-ankle-sprains/

6 0

3 years ago