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AlekseyPX
3 years ago
6

When charges build up on a surface they:

Physics
2 answers:
uysha [10]3 years ago
7 0
I believe the correct answer from the choices listed above is the third option. When charges build up on a surface they, apply a force to charges wandering nearby and <span>create an electric field. </span>

Hope this answers the question. Have a nice day.
Natasha2012 [34]3 years ago
6 0

Answer: A and B

Explanation:

An electric field develops around a charge. The electric field lines represent the region in space where a positive test charge experiences an electric force. A electric force exists between  the two charges.

Thus, when charges build up on a surface, they apply force to charges wandering nearby and also create electric field. So, correct option is Both A and B.

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Compare and contrast speed and velocity
Alex73 [517]
Where speed is distance/time, velocity is displacement/time. 

What this means is that velocity is the length covered in relation to the starting point. 
Speed is just the distance travelled no matter where you began. 

When going around a circular track, you might have a speed value. However, since you get back to the same location at every lap, you have 0 velocity. 

Hope I helped :)
8 0
3 years ago
If Teresa's daughter is my daughters mother who am I to Teresa?
Zolol [24]
Her daughter, nice brainteaser
7 0
3 years ago
Read 2 more answers
Which diagram shows a light ray moving from a more dense into a less dense medium?​
Soloha48 [4]

Answer : it’s A

Explanation:

8 0
3 years ago
How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d
Archy [21]

Answer:

The voltage is   V =   0.993V_b

Explanation:

From the question we are told that

   The time that has passed is  t = \frac{\tau}{2}

 Here \tau is know as the time constant

    The voltage of the  power source is   V_b

Generally the voltage equation for charging a capacitor is mathematically represented as

       V =  V_b  [1 - e^{- \frac{t}{\tau} }]

=>   V =  V_b  [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]

=>   V =  V_b  [1 - e^{- \frac{\tau}{2\tau} }]

=>   V =  V_b  [1 - e^{- \frac{1}{2} }]

=>   V =   0.993V_b    

5 0
3 years ago
A light wave travels through air in equals 1.00 at an angle of 35 degrees what angle does it have when it passes from the air in
lyudmila [28]

Answer: Angle 59 degree

Explanation: Given that the

n1 = 1.0

n2 = 1.5

Øi = 35 degree

From Snell law, which says that

n1/n2 = sinØ1/ sinØ2

Substitute all the parameters into the formula

1/1.5 = sin 35/sinØ2

Cross multiply

Sin Ø2 = 1.5 sin35

SinØ2 = 1.5 × 0.573 = 0.860

Ø2 = sin^-1(0.860)

Ø2 = 59.36 degree

Ø2 = 59 degree ( approximately)

It has angle 59 degree when passing from air to glass

5 0
3 years ago
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