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ololo11 [35]
3 years ago
5

Bits of paper are attracted to an electrified comb or rod, even though they have nonet charge. How is this possible?

Physics
1 answer:
SashulF [63]3 years ago
7 0

Explanation:

An electrified comb is charged comb ( let say by running it through the hair) and when it is brought in the proximity of  pieces of paper, the pieces tend to cling to it. This happens because the charged comb induces an opposite charge in the paper pieces and as opposite charges attract each other, the pieces are clinged.

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Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?
Gemiola [76]

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

A_y=+15 m

\theta=32^{\circ}

We have to find the x-component of A and magnitude of A.

According to question

A_y=\mid A\mid cos\theta

Substitute the values then we get

15=\mid A\mid cos32

\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}

\mid A\mid=17.7m

tan\theta=\frac{perpendicular\;side}{Base}

tan32=\frac{A_x}{A_y}=\frac{A_x}{15}

0.62\times 15=A_x

A_x=9.3

The value of x-component of A is negative because the vector A lie  in second quadrant.

Hence, the x- component of A=-9.3 m

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3 years ago
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If a sealed syringe is plunged into cold water, in which direction will the syringe piston slide? *
Tamiku [17]

Answer:

a

Explanation:

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5 0
2 years ago
The position of a dragonfly that is flying parallel to the ground is given as a function of time by r⃗ =[2.90m+(0.0900m/s2)t2]i^
Arlecino [84]

The solution for this problem is:

 
r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s² 
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v = dr/dt = [0.180t i - 0.0450t² j] m/s² 

tan(-36.0º) = -0.0450t² / 0.180t 
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3 years ago
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umka2103 [35]

Answer:

Radiation

Explanation:

I hope that help's

5 0
3 years ago
If you put 120 volts of electricity through a pickle, the pickle will smoke and start glowing orange-yellow. The light is emitte
Alona [7]

Answer:

2.11eV

Explanation:

We know that speed of light is it's wavelength times frequency.

\therefore f=v/\lambda\\=(3\times10^8m/s)/(589mm\times1m/1\times 10^9nm)\\=5.09\times10^1^4s^-1 \ or \ 5.09\times10^1^4Hz

Planck's constant is 6.626\times 10^3^4Js

The energy gap is calculated by multyplying the light's frequency by planck's constant:

E_c=5.09\times10^1^4s^-^1\times 6.626\times10^-^3^4Js\\\\=3.37\times 10^-^1^9J  \ \ \ \ \ \ #1eV=1.06\times 10^-^1^9J\\\\=2.11eV

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3 years ago
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