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konstantin123 [22]
4 years ago
12

Both objects A and D represent fixed, negatively-charged particles of equal magnitude and Object B represents a fixed, positivel

y-charged particle (equal, but opposite charge from A and D). Object C shows a moving, positively-charged particle. Identify which arrow would correctly show the force of attraction or repulsion on object C caused by the other two objects?
A) 1

B) 2

C) 3

D) 4

Physics
1 answer:
Kipish [7]4 years ago
6 0

Answer:

Option A = 1.

Explanation:

So, in order to solve this question we are given the Important infomation or data or parameters in the question above as;

(1). First, Both objects A and D represent fixed.

(2).  Both objects A and D are negatively-charged particles of equal magnitude.

(3). "Object B represents a fixed, positively-charged particle (equal, but opposite charge from A and D)."

(4). "Object C shows a moving, positively-charged particle."

So, our mission is to determine the arrow that would correctly show the force of attraction or repulsion on object C caused by the other two objects.

We can do that by drawing out the forces of attraction and the resultants. Therefore, CHECK THE ATTACHED FILE/PICTURE FOR THE DRAWINGS.

The forces of attraction due to objects A and B on on object C will be towards themselves. Hence, the resultant is ONE(1).

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Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows
Vlada [557]

Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

Critical angle \theta _c = sin^{-1} (\frac{n}{n_{slab}} )

sin \theta _ c =\frac{n}{n_{slab}}

According to Snell's law of refraction:

n sin \theta _1 = n_{slab}  \ sin \  (90- \theta_c)

At point P ; 90 - \theta _2  \leq \theta _c

\theta _2 = 90 - \theta _c

Therefore:

n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)}  \\ \\ n \ sin \theta_1 = n_{slab}   \sqrt{(1- \frac{n}{n_{slab}} )}

Then maximum value of refractive index  n of the fluid is:

n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }

n = \frac{1.665}{\sqrt{1+ sin^2 \  37} }

n = 1.4266

3 0
3 years ago
Think of a way you could demonstrate elastic force to a younger student. Describe the procedure you would follow and the materia
Soloha48 [4]

Answer:

I feel like to demonstrate you would use an elastic band as the material. You obviously have to put force in order to see how far it stretches. From this you can also find about its resistance and durability

Also you have to make sure the distance between the two hands are equal as you want an accurate result.

7 0
3 years ago
A cube of wood is floating on water. a cube of iron is totally submerged in water. the cubes are equal in volume. which cube has
-BARSIC- [3]
The one that's completely submerged is displacing more water, so the buoyant force on it is greater.
5 0
3 years ago
The human ear canal is about 2.9 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fu
solniwko [45]

The frequency of the human ear canal is 2.92 kHz.

Explanation:

As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is  

Wavelength=4*Length of the ear canal

As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:

wavelength = 4*2.9*10^{-2} =0.116

Then the frequency is determined as

f=c/λ=339/0.116=2922 Hz=2.92 kHz.

So, the frequency of the human ear canal is 2.92 kHz.

4 0
3 years ago
Complete the statement below with all that apply.<br><br> Average speed is what?
iVinArrow [24]

Answer:

a, b, d

Explanation:

6 0
3 years ago
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