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Contact [7]
3 years ago
9

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 10.5 cm, giving it a cha

rge of -22.3 µC.. . (a) Find the magnitude of the electric field just inside the paint layer. (b) Find the magnitude of the electric field just outside the paint layer. (c) Find the magnitude of the electric field 5.00 cm outside the surface of the paint layer.

Physics
2 answers:
soldier1979 [14.2K]3 years ago
8 0
The formula that will be used in this problem is E = q/ 4pi*r^2 z where z is the elctric charge constant equal to 8.854  *10 ^-12. The magnitude using r equal to 0.0525 m and q equal to -22.3 *10^-6 C is equal to -22.3 *10^-6/ 4pi*(0.0525)^2 *8.854  *10 ^-12 or equal to -7.272 *10 ^7. The magnitude 5 cm outside the surface is -22.3 *10^-6<span>/ 4pi*(0.0525+0.05)^2 *8.854  *10 ^-12 equal to -1.908 *10^7. 

</span>
olya-2409 [2.1K]3 years ago
8 0

Since it is hard to write the mathematics equations here so i would like to share it in a picture

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Summarize Newton's three laws of motion
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Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Explanation:

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3 years ago
What are the units, if any, of the particle in a box wavefunction. What does this mean?
SIZIF [17.4K]

Answer:

  • [\psi]= [Length^{-3/2}]
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Explanation:

We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

\int\limits^{x_f}_{x_0} \int\limits^{yf}_{y_0} \int\limits^{z_f}_{z_0} |\psi|^2 \, dz \,  dy \,  dx

must be dimensionless, as represents a probability.

As the differentials has units of length

[dx]=[dy]=[dz]=[Length]

for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:

[\psi]^2 = [Length^{-3}]

taking the square root this gives us :

[\psi] = [Length^{-3/2}]

5 0
3 years ago
A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un
Semenov [28]

Answer:

Explanation:

Given that

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The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

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The magnitude of their distance apart is given as

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r=0.028+0.11

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F=GM1M2/r²

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F=9.58×10^-9N

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F=9.58×10^-9N

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3 years ago
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Answer:Explanation:

Image result for what does a worm and wheel mechanism do to torque and speed​

Like other gear arrangements, a worm drive can reduce rotational speed or transmit higher torque. ... Each full 360 degree turn of a single start worm advances the gear by one tooth. For a multi start worm the gear reduction equals the number of teeth on the gear divided by the number of starts on the worm.

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3 years ago
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So, the river is flowing at a speed 0.975 m/s.

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3 years ago
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